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Geometry practice worksheet answers

Chapter 1 Answers 6. 2. 1.00001, 1.00001 3. 42, 54 4. -64, 128 6. 63.5, 63.75 7. Sample: 2 or 3 8. 51 or 49 9. 6 or 8 10. A or AA 11. D or G 12. Y or A 13. any hexagon 14. circle with 8 equally spaced diameters 15. a 168.75° angle 16. 21 hand17. h = n(n 22 1) 18. 34 19. Sample: The shakes 1 1. 47, 53 5.22, 29 farther out you go, the closer the ratio gets to a number that is approximately 0.618. 20. 0, 1, 1, 2, 3, 5, 8, 13 All rights reserved. Rig 3 Front Front 7. A, C, D 8. C Top 9. D 10. B Right 11. A 1. AC 2. any two of the following: ABD, DBC, CBE, ABE, ECD, ADE, ACE, ACD 3. Points E, B, and D arecollinear. 4. yes 5. yes 6. no 7. no 8. yes 9. no 10. yes 11. yes 12. yes * ) 13. *no ) 14. yes 15. yes 16. *G ) 17. LM 18. PN 19. the empty set 20. KP 21. M 22. *Sample: plane ABD 23. Sample: plane ABC ) 24. AB 25. yes 26. no 27. yes 28. the empty set 29. no 30. yes 31. yes 1. nt 1 Practice * ) 1-3 Practice1-2 Fro Right Practice 1-1 ht 2. Practice 1-4 Fro nt Rig 1. true 6. false 2. false 3. true 4. false 5. false 7. JK, HG 8. EH 9. any three of the * ) * )* ) * )* ) * ) ) * )* ) * )* ) * )* ) * )* ) HG and FK ; JK and EH ; JK and FG ; EJ and FG ; EH * )* ) * )* ) * )* ) * )* ) and FK ; JE and KG ; EH and KG ; JH and KF ; JH and* ) ht following pairs: EF and JH ; EF and GK ; HG and JE ; * 3. 10. planes A and B 11. planes A and C; planes 12. planes A and C 13. planes B and C ) ) ) ) 14. Sample: EG 15. 6 16. EF and ED or EG and ) ) ) ) ) ED 17. FE , FD 18. GF , GD 19. yes Fro B and C nt Rig 1 1 1 1 1 1 20. Sample: Right 4. ht ST Front U 5. 2 2 1 Right Top Front Q Front Front 21. Sample: R Right © Pearson Education, Inc., publishing as Pearson Prentice Hall. GE W P Top Geometry Chapter 1 Right Answers 41 Chapter 1 Answers (continued) 5. Practice 1-5 1. 4 2. 12 3. 20 4. 6 5. 22 6. -10 or 6 7. -1 or 1 8. 3; 4; no 9. 6; 6; yes10. C or -2 11. 15 12. 31 13. 14 14. x = 11 32 ; AB = 31; BC = 31 15. x = 35 23 ; AB = 103; BC = 103 45° Practice 1-6 1. any three of the following: &O, &MOP, &POM, &1 2. &AOB 3. &EOC 4. &DOC 5. 51 6. 90 7. 17 8. 107 9. 141 10. 68 11. &ABD, &DBE, &EBF, &DBF, &FBC 12. &ABF, &DBC 13. &ABE,&EBC 14. x = 5; 8; 21; 13 15. x = 9; 85; 35; 120 90° 6. MN OP B Practice 1-7 7. 1. All rights reserved. A OP MN X B Y A K L 8. 2. 2 1 A Y 1 9. 3. 2 1 2 1 X 2 B Y 10. X 2 ? 2 2 4. C 11. X Z 42 Answers Geometry Chapter 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. X Chapter 1Answers (continued) 12. Practice 1-9 1. 792 in.2 2. 3240 in.2 3. 2.4 m2 4. 32p 5. 16p 6. 7.8p 7. 26 cm; 42 cm2 8. 46 cm; 42 cm2 9. 29 in.; 42 in.2 10. 40 ft; 51 ft2 11. 40 m; 99 m2 2 12. 40 m; 91 m 13. 68; 285 14.3 26; 22 15. 30; 44 16. 156.25p 17. 10,000p 18. p16 19. 48; 28 20. 36 21. 26; 13 2 ? X WReteaching 1-1 1. 13. , All rights reserved. X Z 14. true 19. false 15. false 20. true 2. 16. false 17. true 18. true , Practice 1-8 1.–5. 3. y 6 A , C 4 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 6 D 4 E 4. 2 2 6 x 4 2 B , 4 5. 6 6. 5"2 7.1 7. 2"17 8.2 8. 12 9. 8 10. 12 11. "26 5.1 12. (5, 5) 13.(21 , 1) 14. (10 12 , -5) 15. (-2 12 , 6) 16. (-0.3, 3.4) 5 7 17. (2 8 , -8 ) 18. (5, -2) 19. (4, 10) 20. (-3, 4) 21. yes; AB = BC = CD = DA = 6 22. "401 20.025 23. y S , 6. , Reteaching 1-2 1a. 6 Q 4 2 P 24. 24.7 R 2 4 6 Fro nt Rig ht 8x 25. (3.5, 3) Geometry Chapter 1 Answers 43 Chapter 1 Answers 1b.(continued) 6. Front Top Right Fro 2a. nt Rig ht Reteaching 1-3 1.–6. Check students’ work. Fro nt h Rig 7. t 6 y S 2 Front Top 6 4 2 O P2 Right 4 3a. R Fro nt Rig 2 4 6x T 6 8. Sample: R, Q, T 9. P, R, S * ) * ) 10. Sample: RS and QT ht Front Top Right 4a. Samples: * ) * ) * ) * ) * ) * ) 1. AB and EF 2. ABand DH 3. AC and AE 4. plane ABD and plane EFH 5. plane CDH and plane BFD 6. plane CDH, plane ACG, and plane ABD * ) * ) 7. EG and BF 8. Fro nt h Rig 9. t 4b. Front Top Right 10. 5. Front Top Answers Right Geometry Chapter 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall.Reteaching 1-4 3b. 44 All rights reserved. 4 Q 2b. Chapter 1 Answers (continued) 4. 11. D 12. E F 5. T 13. All rights reserved. U V 6. J Reteaching 1-5 1. 40 km 6. Ryan 2. 25 km 3. 30 km 4. 30 km 5. 60 km 7. 100 km; 25 + 15 + 30 + 30 = 100 Reteaching 1-6 K L Reteaching 1-8 1. 6 © Pearson Education,Inc., publishing as Pearson Prentice Hall. 1a. Each angle is 108°. 1b. obtuse 2a. Each angle is 72°. 2b. acute 3a. Each angle is 108°. 3b. obtuse 4. 90°, right 5. 55°, acute 6. 110°, obtuse 7. 50°, acute 8. 90°, right 9. 100°, obtuse Reteaching 1-7 X 4 2 6 Samples: y 4 1. 2 2 2 4 C 4 6x Z Y 6 A B 2. Q 2. XY10.6, YZ 4.1, XZ 7.2 3. 4.1 + 7.2 = 11.3, 11.3 10.6 4. 12.2 5. 8.1 6. 9.9 7. 11.4 8. 9.2 9. 8.5 10. 18.0 11. 8.5 12. 17 Reteaching 1-9 R 3. S X 1a. 1 by 60, 2 by 30, 3 by 20, 4 by 15, 5 by 12, and 6 by 10; Check students’ drawings. 1b. 122, 64, 46, 38, 34, and 32 2a. 1 by 36, 2 by 18, 3 by 12, 4 by 9 and 6 by 6;Check students’ drawings. 2b. 1 by 36 3. 3 cm by 6 cm 4. Sample: diameter 8 units; radius 4 units 4 Y Z Geometry Chapter 1 Answers 45 Chapter 1 Answers 5. Sample: 50 units2 7. They are close. (continued) 6. Sample: 50.3 units2 6. Enrichment 1-1 Top 2. 16 4. Round 1 2 3 4 5 6 Round of 64 32 16 84 2 Games in Round 32 16 8 4 2 1 Total Games Played 32 48 56 60 62 63 Bottom Left side Enrichment 1-2 Back 7. Top 5. The number in the fourth column is equal to the sum of the numbers in the third column up to the given row. 6. 16 7. 1 + 2 + 22 + c + 2 n = 2 n + 1 - 1 8. 127 9. 3 1. 4 Right side Front3. 8 Right side Front Bottom Left side Back Right side Front Bottom Left side Back All rights reserved. 1. 32; 32; 32 8. 2. 7 3. Top nt Rig ht 4. 9. Top Right side Front Right side Front Bottom Left side Back Top Bottom Left side Back 5. Top Right side Bottom Left side 46 Front Back Answers 10. V = 9units3; S.A. = 32 units2 11. V = 9 units3; 2 3 S.A. = 34 units 12. V = 9 units ; S.A. = 34 units2 13. V = 7 units3; S.A. = 28 units2 14. V = 11 units3; S.A. = 34 units2 15. V = 14 units3; S.A. = 44 units2 16. Sample: The foundation drawings; you can find the sum of the numbers on each square. 17. Sample:When there are no valleys in the cube arrangement, you can count the total number of squares used in all six views of an orthographic projection to find the surface area. Geometry Chapter 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Fro Chapter 1 Answers (continued) 11.Enrichment 1-3 1. No; they will be parallel. 2. No; they will be skew. 3. S1 and S3; T and B; S2 and S4 4. S3 5. S4 6. line l or L3 7. line b 8. line n or L2 9. line a 10. 1; S3 11. 0 12. 0 13. 1; S3 14. S1 and S4 15. T and S1 16. S2 17. S3 18. Check students’ work. 19. Check students’ work. Enrichment 1-4 Allrights reserved. 1. north–south 2. east–west 3. east–west 4. north–south 5. always 6. never 7. never 8. always 9. never 10. 1000 ft 11. Sample: If one airplane is flying east–west and one is flying north–south, then one was out of its proper altitude range. Sample: If both are flying east–west at the samecruising altitude, then the air traffic controller instructed one of the airplanes incorrectly. 12. First Avenue, Second Avenue, or Third Avenue 13. Park Street, First Avenue, Second Avenue, or Grove Street 14. Second Avenue or First Avenue 15. No; the streets are parallel. 16. Carter Street, Oak Street, HillStreet, Pine Street, and Crosstown Road Enrichment 1-5 1. 1 mi 2. 9 mi 3. 5 mi 4. ABDE, 12 mi; ABCFE, 14 mi 5. F; more shorter connections 6. I; only one long connection 7. link E with I; it is short and would or Enrichment 1-8 1.–7., 10.–14., 16.–19. C B D E F A H G 8. "32 9. "32; Check students’ work.20. triangle 15. trapezoid Enrichment 1-9 increase the number of connections to I. 1.–4. Sample: © Pearson Education, Inc., publishing as Pearson Prentice Hall. Enrichment 1-6 1. -4 2. 1 12 3. 21 1 1 7. 4 2 units 8. 5 2 units 4. III 5. II 9. 1 unit 6. I Enrichment 1-7 A 1.–6. B B A F F I E C D C E students’work. Early mathematicians’ beliefs were incorrect as stated in Exercise 5. D 7.–8. A B B 5. Answers may vary. 6.–12. Check students’ work. 13. They are all the same, about 3.142; p. 14. Check Chapter Project A F F • Table: 4; 8; 16 • 32; 64; multiply by 2 I E C C D Activity 1: Paper Folding E D 9.hexagon ABBAFF and quadrilateral ICDE 10. hexagon CDEEDC and quadrilateral AFIB Activity 2: Creating Check students’ work. Activity 3: Writing Check students’ work. Activity 4: Researching Check students’ work. Geometry Chapter 1 Answers 47 Chapter 1 Answers (continued) ✔ Checkpoint Quiz 11 1. 91 , 27 2. 49, 56 3. G 4. Any three points are 5. intersecting 6. skew 7. parallel coplanar. 8. 9. E, G, O, and F* ) 10. Sample: A, D, F, and B 11a. O 11b. EF 11c. O 11d. G 12. Yes; they are vertical angles. 13. No; there are no markings. 14. Yes; they are adjacent and m&AOD + m&DOC = 180. 15. No;you cannot conclude this from the diagram. 16a. 11 16b. 22 16c. 40 17. 56 18. 56 19. 34 20. 90 21. 146 22. 112 23. 63°; acute 24. 102°; obtuse 25. 90°; right 26. 9. ht Rig nt 2 1 1 1 A All rights reserved. Fro 1 1 27. 1 P Q ✔ Checkpoint Quiz 2 1. 25 2. 28. A B 29. R T P Q A 1 – ? lP 2 P 4. 10 30. 34. 37. 39.5. (3, 0) 6. B 7. 16p ft 8. 64p ft2 Chapter Test, Form B Chapter Test, Form A 1 1 -21 ; -16 , 32. 1. Add 5; 32, 37. 8.1 31. 8.5 32. 19.7 33. (4, -11) 52 ft; 169 ft2 35. 38 cm; 70 cm2 36. 32 in.; 42 in.2 1 2 2 p ft; 4 p ft 38. 3p cm; 2.25p cm 22p in.; 121p in.2 2. Multiply by 3. Add 4, then 5, then 6, and so on; 29,37. 4. Add another side of a hexagon; check students’ work. 5. The relationship (x - 1)(x + 1) = x2 - 1 is true for all whole numbers x. No counterexample exists. 6. Sample: 6 ft 1. Multiply by 3; 81, 243. 2. Add 11, add 9, add 11, add 9, and so on; 44, 55. 3. Insert one more dot; a circle with three dots, acircle with four dots 4. The relationship (x 2 1) 1 (x 1 1) = 2 x is true for all whole numbers x. No counterexample exists. 5. Sample: 6. Sample: 2 cm 6 ft 10 cm 7 ft 4 cm 7. Sample: 3m 10 m 8. Possible answers: A, O, and B; E, O, and F; C, O, and D 48 8 in. 22 in. Answers 2 cm 7. M, E, B, and D 8. C, M,and O or M, B, and E 9. Sample: A, F, O, and D 10a. M 10b. M 10c. F 10d. O 11. &AOB 12. &DOE 13. &COD 14. &AOE and &BOC or &AOB and &COE 15a. 5 15b. 9 15c. 48 16. 25 17. 25 Geometry Chapter 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3. Chapter 1 Answers 18. 13019. 180 20. 140º; obtuse (continued) 21. 63º; acute 22. P TASK 2: Scoring Guide a. Sample: 23. A 2 Student gives explanations, descriptions, and a drawing that are generally clear but may contain a few errors. 1 Student makes significant errors in explanations and descriptions. 0 Student makes little orno attempt. B C P A All rights reserved. 24. G D P b. No; any three points are coplanar. c. Sample: A, B, and C 25. A B 26. 5 27. 17.3 28. (9, 18) 29. 58 in.; 168 in.2 2 30. 38 cm; 78 cm 31. 8p in.; 16p in.2 2 32. 30p cm; 225p cm © Pearson Education, Inc., publishing as Pearson Prentice Hall. B E 3 Studentgives an accurate diagram, correct answers, and a valid explanation. 2 Student draws a diagram that contains some errors but gives answers to parts b and c that are correct. 1 Student draws a diagram that contains significant errors and gives answers to parts b and c that are incorrect. 0 Student makeslittle or no attempt. TASK 3: Scoring Guide Sample: P Alternative Assessment, Form C TASK 1: Scoring Guide Samples: a. Exterior rays form a right angle for each figure. The total number of rays doubles with each successive figure. Two parallel segments connect the innermost rays. b. Other answersare possible. For example, instead of considering the total number of rays (2, 4, 8, 16), consider the rays in the interior of the right angle: 0, 2, 6, . . . In that case, you can argue that 12 comes next, giving a total of 14 rays, not 16, in figure 4. c. same as in part b, except with 32 rays (If considering the raysin the interior of the right angle, Figure 5 will have 20 rays.) C (90 + mlPQR)° B A R Q 3 Student constructs the figures correctly. 2 Student constructs the figures, but the constructions may contain some minor errors. 1 Student misses some key ideas, resulting in significant errors in the constructions. 0Student makes little or no attempt. 3 Student gives a correct explanation, clear descriptions, and an accurate drawing. Geometry Chapter 1 Answers 49 Chapter 1 Answers (continued) 21. Sample: TASK 4: Scoring Guide Foundation drawing: 1 1 2 2 2 2 1 E F All rights reserved. Orthographic drawing: 22.Sample: Front Top Right 3 Student draws completely accurate figures. 2 Student draws generally accurate figures, but these may contain a minor error. 1 Student draws figures containing significant errors. 0 Student makes little or no attempt. 45° 23. (-4, -2) 24. (0, -3) 25. "29 Cumulative Review 2. F 3. D4. G 5. D 6. G 7. C 9. C 10. J 11. B 12. G 13. B 14. Sample: 1, 2, 3, 4, 5, 6, c and 1, 2, 3, 1, 2, 3, 1, c The first is the natural numbers. In the second, the numbers 1, 2, and 3 repeat. 15. Yes; the markings show they are congruent. 16. No; there are no markings. 17. Yes; you can conclude the angles aresupplementary from the diagram. 18. sometimes 19. never 20. Not acceptable; too general; Sample: a ruler is a tool used for measuring lengths and is marked showing uniform units. 50 Answers © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. A 8. J Geometry Chapter 1 Chapter 2Answers Practice 2-1 1. Sample: It is 12:00 noon on a rainy day. 2. Sample: The car will not start because of a dead battery. 3. Sample: 6 4. If you are strong, then you drink milk. 5. If a rectangle is a square, then it has four sides the same length. 6. If you are tired, then you did not sleep. 7. If x = 26, thenx - 4 = 22; true. 8. If x 0, then ∆x∆ 0; true. 9. If m is positive, then m2 is positive; true. 10. If 2y - 1 = 5, then y = 3; true. 11. If x 0, then point A is in the first quadrant; false; (4, -5) is in the fourth quadrant. 12. If lines are parallel, then their slopes are equal; true. 13. If you have a sibling, then you are a twin;false; a brother or sister born on a different day. All rights reserved. 14. siblings twin 15. Hypothesis: If you like to shop; conclusion: Visit Pigeon Forge outlets in Tennessee. 16. Pigeon Forge outlets are good for shopping. 17. If you visit Pigeon Forge outlets, then you like to shop. 18. It is not necessarilytrue. People may go to Pigeon Forge outlets because the people they are with want to go there. 19. Drinking Sustain makes you train harder and run faster. 20. If you drink Sustain, then you will train harder and run faster. 21. If you train harder and run faster, then you drink Sustain. © Pearson Education,Inc., publishing as Pearson Prentice Hall. Practice 2-2 1. Two angles have the same measure if and only if they are congruent. 2. 2x - 5 = 11 if and only if x = 8. 3. The converse, “If ∆n∆ = 17, then n = 17,” is not true. 4. A figure has eight sides if and only if it is an octagon. 5. If a whole number is a multipleof 5, then its last digit is either 0 or 5. If a whole number has a last digit of 0 or 5, then it is a multiple of 5. 6. If two lines are perpendicular, then the lines form four right angles. If two lines form four right angles, then the lines are perpendicular. 7. If you live in Texas, then you live in the largest state in thecontiguous United States. If you live in the largest state in the contiguous United States, then you live in Texas. 8. Sample: Other vehicles, such as trucks, fit this description. 9. Sample: Other objects, such as spheres, are round. 10. Sample: This is not specific enough; many numbers in a set could fit thisdescription. 11. Sample: Baseball also fits this definition. 12. Sample: Pleasing, smooth, and rigid all are too vague. 13. yes 14. no 15. yes Practice 2-3 1. &A and &B are complementary. 2. Football practice is canceled for Monday. 3. DEF is a right triangle. 4. If you liked the movie, then you enjoyedyourself. 5. If two lines are not parallel, then they intersect at a point. 6. If you vacation at the beach, then you like Florida. 7. not possible 8. Tamika lives in Nebraska. Geometry Chapter 2 9. not possible 10. It is not freezing outside. 11. Shannon lives in the smallest state in the United States. 12. OnThursday, the track team warms up by jogging 2 miles. Practice 2-4 1. UT = MN 2. m&QWR = 30 3. SB = MN 4. y = 51 5. JL 6. Given; Addition Property of Equality; Division Property of Equality 7. Given; Addition; Subtraction Property of Equality; Multiplication Property of Equality; Division Property ofEquality 8. Substitution 9. Substitution 10. Transitive Property of Equality 11. Symmetric Property of Congruence 12. Transitive Property of Congruence 13. Definition of Complementary Angles; 90, Substitution; 3x, Simplify; 3x, 84, Subtraction Property of Equality; 28, Division Property of Equality 14.Given; (2x - 4), Substitution; 6x - 12, Distributive Property; -x, Subtraction Property of Equality; 12, Multiplication Property of Equality Practice 2-5 1. 30 2. 15 3. 20 4. 6 5. 16 6. 9 7. m&A = 135, m&B = 45 8. m&A = 36, m&B = 144 9. m&A = 75, m&B = 15 10. m&A = 10, m&B = 80 11. m&PMO = 55; m&PMQ= 125; m&QMN = 55 12. m&BOD = m&COE = 90; m&BOC = m&COD = 45; m&AOB = m&DOE = 45 13. m&BWC = m&CWD, m&AWB + m&BWC = 180; m&CWD + m&DWA = 180; m&AWB = m&AWD Reteaching 2-1 1.–3. Check students’ work. 4. If you hear thunder, then you see lightning; statement:false; converse: false. 5. If your pants are jeans, then they are blue; statement: false; converse: false. 6. If you are eating a tangerine, then you are eating an orange fruit; statement: false; converse: true. 7. If a number is an integer, then it is a whole number; statement: true; converse: false. 8. If a trianglehas one angle greater than 90°, then it is an obtuse triangle; statement: true; converse: true. 9. If n2 = 64, then n = 8; statement: true; converse: false. 10. If you got an A for the quarter, then you got an A on the first test; statement: false; converse: false. 11. If a figure has four sides, then it is a square;statement: true; converse: false. 12. If x = 144, then "x = 12; statement: true; converse: true. Reteaching 2-2 1. If n = 15 or n = -15, then ∆n∆ = 15. If ∆n∆ = 15, then n = 15 or n = -15. 2. If two segments are congruent, then they have the same measure. If two segments have the same measure, then theyare congruent. 3. If you live in California, then you live in the most populated state in the United States. If you live in the most populated state in the United States, then you live in California. 4. If an integer is a multiple of 10, then its last digit is 0. If an integer’s last digit is 0, then it is a multiple of 10.Answers 29 5. No; counterexamples may vary. Sample: A giraffe is a large animal. 6. Yes; two planes intersect if and only if they form a line. 7. Yes; a number is an even number if and only if it ends in 0, 2, 4, 6, or 8. 8. Yes; a triangle is a threesided figure if and only if the angle measures sum to 180°.Reteaching 2-3 1. Law of Detachment; the police officer will give Darlene a ticket. 2. Law of Syllogism; if two planes do not have any points in common, then they are parallel. 3. Law of Detachment; Landon has a broken arm. 4. Not possible; a conclusion cannot be drawn from a conditional and itsconfirmed conclusion. (Brad may live in Peoria, Illinois.) 5. Law of Detachment; the circumference is p. Reteaching 2-4 1. Addition Property of Equality 2. Symmetric Property of Equality 3. Distributive Property 4. Subtraction Property of Equality 5. Addition Property of Equality 6. Multiplication Property ofEquality 7. Substitution 8. Transitive Property of Equality 9. 4, 5, 1, 3, 2 Reteaching 2-5 1.–7. 5, 2, 4, 1, 6, 3, 7 or 5, 2, 1, 4, 6, 3, 7 8. Sample: m&10 + m&6 = 180 by the Angle Addition Postulate. m&6 + m&7 + m&8 = 180 by the Angle Addition Postulate. m&10 + m&6 = m&6 + m&7 + m&8 by the TransitiveProperty of Equality. Subtract from both sides, and you get m&10 = m&7 + m&8. Because &10 is a right angle, m&7 + m&8 = 90. It is given that m&7 = m&8. By Substitution, 2m&8 = 90. Divide both sides by 2, and m&8 = 45. 9. Sample: Because &9 and &10 form a straight angle, m&9 + m&10 = 180 bythe Angle Addition Postulate. By the definition of supplementary angles, &9 and &10 are supplementary. 10. Sample: Because &10 and &6 form a straight angle, m&10 + m&6 = 180 by the Angle Addition Postulate. &10 is a right angle, so 90 + m&6 = 180. Subtract 90 from both sides, and you get m&6 =90. So m&6 is a right angle by the definition of right angles. Enrichment 2-1 1. point 2. postulate 3. parallel 4. perpendicular 5. perimeter 6. plane 7. protractor 8. pentagon 9. Sample: W, L 10. LXCBR 11. Sample: If LWHJP stands for plane, then LHXHWWPW must be parallel because of the repetition ofthe letter W. 12. Check students’ work. Enrichment 2-2 1. If a kitchen is well-designed, then the distance in a straight line from the center point of the sink, refrigerator, or stove to another is between 4 feet and 9 feet. If the distance in a straight line from the center point of the sink, refrigerator, or stove toanother is between 4 feet and 9 feet, then the kitchen 30 Answers is well-designed. 2. The statement can be written as a biconditional because both the conditional and its converse are true. 3. A kitchen is considered to be well-designed if and only if the distance in a straight line from the front center pointof one of these three objects to another is between 4 feet and 9 feet. Additionally, the sum of the three triangle legs should be less than or equal to 26 feet. 4. The definition is a good definition. The terms used are commonly understood. The definition is precise. The definition is reversible: If the distancesin a straight line between the front center points of the sink, refrigerator, and stove are between 4 feet and 9 feet, and the sum of these distances is less than or equal to 26 feet, then the kitchen is considered to be well-designed. 5. Yes; the distance from the sink to the refrigerator is 6.5 feet, the distancefrom the refrigerator to the stove is 6.25 feet, and the distance from the stove to the sink is 9 feet. The sum of these distances is less than 26 feet. 6. Check students’ work. Enrichment 2-3 1. p: one is a student at Richmond High School; q: one takes geometry class; r: one studies logic. 2. If someone is astudent at Richmond High School, then he or she takes geometry class. 3. If someone takes geometry class, then he or she studies logic. 4. All students at Richmond High School study logic. 5. Logic can be studied in classes other than geometry, such as algebra or trigonometry. 6. a; p: a figure is asquare; q: a figure is a polygon; r: a figure is closed; if p S q and q S r then p S r. 7. b; p: a figure is a square; q: a figure is a rectangle; r: a figure has four right angles; if p S q and q S r, then p S r. 8. All mathematicians enjoy puzzles. 9. Triangles have an angle sum of 180. 10. Mt. McKinley is 20,320 fttall. 11. Nolan Ryan recorded 5714 strikeouts over his career. 12. Trapezoids are four-sided figures. Enrichment 2-4 D 8 B T R S T I T I O N O N B S I V I D I T I T I O N M U L I O N 11 I C L C P I I R E T T T V A C I L S L E I N E V I T I M T N O I T C A R T M I O I T I D S N A U 7 Y A 9 10 U U 1 6 5 R 24 C O U T N T I G I O R N U T R O I A N P N D 3 A 12 13 E Q N U U C A B E L I T Y Enrichment 2-5 1. They are complementary angles; if the exterior sides of two adjacent acute angles are perpendicular, then the angles are complementary. 2. &ECD 3. They are equal; Geometry Chapter 2 All rightsreserved. (continued) © Pearson Education, Inc., publishing as Pearson Prentice Hall. Chapter 2 Answers Chapter 2 Answers (continued) because &ACB and &DCB have equal measures, you can use the definition of supplementary angles and the subtraction property to show that they are equal. 4. 90 5.135 6. 45 7. NO bisects TP; a segment that is perpendicular to another segment at its midpoint is the perpendicular bisector of the segment. Chapter Project Check students’ work. All rights reserved. ✔ Checkpoint Quiz 1 1. hypothesis: x + 4 = 10; conclusion: x = 6 2. hypothesis: if you want to get goodgrades in school; conclusion: you must study hard 3. Sample: It could be May. 4. Sample: Corn is a vegetable. 5. No; the lines must be in the same plane. 6. yes 7. not possible 8. X, Y, and Z are coplanar. 9. If you ran a good race, then your coach is happy. 10. If the car is old, then it is not efficient. ✔Checkpoint Quiz 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. 4x = 13 2. &TQM <S 3. Symmetric Property 4. Substitution Property 5. Division Property 6. Addition Property 7. &1 &3, &2 &4; Vertical Angles Theorem 8. &2 &3, &1 &4; Linear Pair and Transitive Property of Equality 9.&1 &4, &2 &3; Subtraction Property of Equality 10. &1 &2, &3 &4; Vertical Angles Theorem Chapter Test, Form A 1a. If a polygon has three sides, then it is a triangle. 1b. true 2a. If George lives in the United States, then he lives in Texas. 2b. false 3a. If two angles are congruent, then they are verticalangles. 3b. false 4. Addition Property of Equality 5. Transitive Property of Equality 6. Subtraction Property of Equality 7. Division Property of Equality 8. Reflexive Property of Equality 9a. 90 9b. 58 9c. 148 9d. 90 9e. 122 9f. 148 10. 17 11. 6 12. 14 13. 11 14. not possible 15. We win. 16. If the bus is late,then we will receive a tardy penalty. 17. The sum of the measures of &A and &B is 90. 18. If a quadrilateral is a rectangle, then it has four right angles. If a quadrilateral has four right angles, then it is a rectangle. 19. A rhombus has four congruent sides. 20. good definition 21. good definition 22. A bat haswings. 23a. Distributive Property 23b. Addition Property of Equality 23c. Division Property of Equality 24. 25 25. 19 26. 17 27. Answers may vary. Sample: (1, 2) 28. Answers may vary. Sample: (3, -4) Chapter Test, Form B 1a. If a polygon has five sides, then it is a pentagon. 1b. true 2a. If Mary lives inMinnesota, then she lives in Geometry Chapter 2 Minneapolis. 2b. false 3. Reflexive Property of Equality 4. Substitution Property of Equality 5. Division Property of Equality 6. Distributive Property 7a. 38 7b. 128 7c. 26 7d. 154 8. 50 9. 13.5 10. 4 11. Martina is quick. 12. If I don’t wear sunscreen whileswimming, then I’ll be in pain. 13. If a quadrilateral is a parallelogram, then it has two pairs of opposite sides parallel. If a quadrilateral has two pairs of opposite sides parallel, then it is a parallelogram. 14. A rectangle has four right angles. 15. good definition 16. An octopus has eight legs. 17a. DivisionProperty of Equality 17b. Subtraction Property of Equality 17c. Division Property of Equality 18. 17 19. 28 20. Answers may vary. There are two possibilities. Samples: (4, -2) or (-2, 5) Alternative Assessment, Form C TASK 1: Scoring Guide a. Sample: If it is raining, then the lawn is wet. b. Sample: If thetemperature is below 32°F, then the temperature is below freezing. 3 Student gives conditionals that meet the criteria and provides clear and accurate explanations of why the conditionals meet the criteria. 2 Student gives examples and explanations that are generally clear but may contain errors. 1Student makes significant errors in examples or explanations. 0 Student makes little or no attempt. TASK 2: Scoring Guide Sample: The definition cannot be written as a biconditional. A school is a place where people are educated. 3 Student gives an explanation and a definition that are accurate, clear,and complete. 2 Student gives an explanation and a definition that are generally clear but may contain errors. 1 Student makes significant errors in explanation or definition. 0 Student makes little or no attempt. TASK 3: Scoring Guide a. Sample: If two angles have sides that are opposite rays, then they arevertical angles. &CEF and &GED are angles whose sides are opposite rays. Therefore, &CEF and &GED are vertical angles. b. Sample: &CEF and &GED are vertical angles. All vertical angles are congruent. Therefore, &CEF &GED. 3 Student gives accurate and complete examples and explanations ofthe Laws of Detachment and Syllogism. 2 Student gives examples and explanations that are generally correct but may be unclear. 1 Student gives examples or explanations that contain major errors. 0 Student makes little or no attempt. Answers 31 Chapter 2 Answers (continued) Cumulative ReviewTASK 4: Scoring Guide Sample: 1. b 2. a 3. C 4. G 5. A 6. H 7. C 8. F 9. D 10. G 11. C 12. J 13. C 14. 17 15. If a polygon has three sides, then it is a triangle; true. 16. 196p m2 17. Sample: Two y 3 2 R T P 1 V 1 2 1 1 S 2 3 x supplementary angles are not necessarily a linear pair. * ) 18. BC 19.Substitution Property of Equality 20. (-1, 2) 21. (-2, 2) 22. "145 23. 4 24. 60 m2 25. 380.13 cm2 26. Sample: A midpoint must be on the segment. Any point that lies on the perpendicular bisector of a segment is equidistant from the endpoints. a. &TPR and &RPV; &RPV and &VPS; &VPS and All rightsreserved. &SPT; &SPT and &TPR; &RPT and &VPS; &TPS and &RPV b. &RPT and &VPS; &TPS and &RPV c. &RPT is a right angle because the lines are perpendicular, so m&RPT = 90. Therefore, if ray PY bisects &RPT, m&YPR = 45. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3Student draws a diagram and gives responses that indicate a thorough understanding of the definitions involved. Student shows logic in part c that is complete and accurate. 2 Student draws a diagram and gives responses that, while basically accurate, contain some minor flaws, inaccuracies, oromissions. 1 Student draws a diagram or gives responses that contain significant inaccuracies or omissions. Student displays flaws in logical argument in part c. 0 Student makes little or no attempt. 32 Answers Geometry Chapter 2 All rights reserved. Chapter 3 Answers Practice 3-1 Practice 3-4 1.corresponding angles 2. alternate interior angles 3. same-side interior angles 4. alternate interior angles 5. same-side interior angles 6. corresponding angles 7. 1 and 5, 2 and 6, 3 and 8, 4 and 7 8. 4 and 6, 3 and 5 9. 4 and 5, 3 and 6 10. m1 = 100, alternate interior angles; m2 = 100, corresponding anglesor vertical angles 11. m1 = 75, 1. 125 2. 69 3. 143 4. 129 5. 140 6. 136 7. x = 35; y = 145; z = 25 8. a = 55; b = 97; c = 83 9. v = 118; w = 37; t = 62 10. 50 11. 88 12. m3 = 22; m4 = 22; m5 = 88 13. 57.1 14. 136 15. m1 = 33; m2 = 52 16. isosceles 17. obtuse scalene 18. right scalene 19. obtuse isosceles20. equiangular equilateral alternate interior angles; m2 = 75, vertical angles or corresponding angles 12. m1 = 135, corresponding angles; m2 = 135, vertical angles 13. x = 103; 77°, 103° 14. x = 24; 12°, 168° 15. x = 30; 85°, 85° 16a. Alternate Interior Angles Theorem 16b. Vertical angles are congruent.16c. Transitive Property of Congruence Practice 3-2 * ) * ) 1a. same-side interior 1b. QR 1c. TS 1d. * same-side interior 1e. Same-Side Interior Angles ) 1f. TS 1g. 3-5 2. l and m, Converse of Same-Side Interior Angles Theorem 3. none 4. BC and AD , Converse of Same-Side Interior Angles Theorem 5.RT and HU , Converse of Corresponding Angles Postulate 6. BH and CI, Converse of Corresponding Angles Postulate 7. a and b, Converse of Same-Side Interior Angles Theorem 8. 43 9. 90 10. 38 11. 100 12. 70 13. 48 Practice 3-5 1. x = 120; y = 60 2. n = 5137 3. a = 108; b = 72 4. 109 5. 133 6. 129 7.129 8. 47 9. 127 10. 30 11. 150 12. 6 13. 5 14. 8 15. BEDC 16. FAE 17. FAE and BAE 18. ABCDE 19. 20 Practice 3-6 1. y = 31 x - 7 4. y 9. 1. True. Every avenue will be parallel to Founders Avenue, and © Pearson Education, Inc., publishing as Pearson Prentice Hall. = 16 x - 32 y 6 4 2 y 4x 1 2 4 6 x y 1x 2 3 2 4 6 x –6–4–2 –2 –4 –6 y 1 x 5 4 x 4 2 –6–4 –2 –2 –4 –6 2 4 6 x –6–4–2 –4 –6 y 15. 6 4 2 2 4 6 6 y 2x 7 4 2 –6–4 –2 –2 –4 –6 y yx1 12. y 6 4 2 –6–4–2 –2 –4 –6 13. 6. y = 45 x - 2 10. –6 11. 3. y = 7x - 18 8. y = -x + 6 y 6 4 2 –6–4–2 –2 l2 5. y 3 7. y = 4x - 13 Practice 3-3 therefore every avenuewill be perpendicular to Center City Boulevard, and therefore every avenue will be perpendicular to any street that is parallel to Center City Boulevard. 2. Not necessarily true. No information has been given about the spacing of the streets. 3. True. The fact that one intersection is perpendicular, plus thefact that every street belongs to one of two groups of parallel streets, is enough to guarantee that all intersections are perpendicular. 4. True. Opposite sides of each block must be of the same type (avenue or boulevard), and adjacent sides must be of opposite type. 5. Not necessarily true. If there aremore than three avenues and more than three boulevards, there will be some blocks bordered by neither Center City Boulevard nor Founders Avenue. 6. a ⊥ e 7. a || e 8. a || e 9. a ⊥ e 10. a ⊥ e 11. a || e 12. If the number of ⊥ statements is even, then 1 || n. If it is odd, then 1 ⊥ n. 13. The proof caninstead use alternate interior angles or alternate exterior angles (if the angles are congruent, the lines are parallel) or same-side interior or same-side exterior angles (if the angles are supplementary, the lines are parallel). 14. It is possible. 2. y = -2x + 12 1 = -2 x - 14. 2 4 6 x y 6 y 2x 3 4 2 –6–4–2 –2 y 16.y 3x 5 2 4 6 x y 2 6 y x2 3 4 4 x 2 4 6 –6 –2 –2 –4 –6 2 4 6 x l1 l3 Geometry Chapter 3 Answers 39 Chapter 3 Answers 2 4 6 x –6–4–2 20. y 6 4 2 x 2 –6–4 2 4 6 x –2 –4 –6 21. 22. y –4 –6 –6 y 23. y 6 4 2 y – 3x 2 2 –6–4–2 –2 –4 –6 yx 2 4 6 x –4 –6 24. 2 4 6 x x 2.5 2 4 6 x –6–4–2 –2 –4 –6 25. y = –3x+ 13 26. y = x + 4 27. y = 12 x - 3 1 1 28. y = -2 x - 2 29. y = 2x + 4 30. y = 13 x + 4 31. y = -51 x - 65 32. y = -6x + 45 33. x = 2; y = -11 34. x = 0; y = 2 35. x = -4; y = -4 36. x = -1; y = 8 37. y 6 4 (4, 0) 2 38. y (–2, 0) 6 4 2 –6–4–2 2 6 x –2 –4 –6 –8 –10 –12 (0, 12) –6 –4–2 –2 (0, 6) (9, 0) x 2 4 6 8 –4 –6 –2–2 2 4 6 x –4 –6 45a. m = $0.10 45b. the amount of money the worker is paid for each box loaded onto the truck 45c. b = $3.90 45d. the base amount the worker is paid per hour 46. y = -25 x + 8 Practice 3-7 y 6 4 2 2 4 6 8 x y 6 4 2 6 4 (–3, 0) 2 2 4 6 x –6–4–2 y 5 44. 12 (0, 12) 10 6 4 2 2 4 6 x –6–4–2 –2 –6–4–2 –2 –4 –6 y 6 4 –6–4–2 –2 –4 –6 6 4 2 y 6 4 2 (0, 1) (8, 0) 2 4 6 x –6–4–2 –2 –4 (0, –4) –6 43. y y 2x 42. y 6 4 2 (4, 0) 2 6x –6–4–2 –2 1 –4 y 2 x 3 –6 –4 –6 19. 41. y 6 4 2 –6 –4 –2 2 4 6 x –2 –4 –6 (0, –1) 1. neither; 3 2 13, 3 ? 13 2 -1 2. perpendicular; ? -2 = -1 3. parallel; -32 = -23 4. parallel; -1 = -1 5. perpendicular; y = 2 is a horizontal line, x = 0 is a vertical line 6. parallel; -21 = -21 7. neither; 1 2 18, 1 ? 18 2 -1 8. parallel; -32 = -23 9. perpendicular; -1 ? 1 = -1 10. neither; 12 2 7 5 1 5 7 2 -3, 2 ? –3 2 -1 11. neither; -3 2 -12 , -23 ? -12 2 -1 1 1 12. neither; 6 2 -5, 6 ? -5 2 -1 13. neither; 92 2 4, 92? 4 2 -1 14. parallel; 21 = 12 15. y = 32 x 4 16. y = -3 x + 24 y 17. y = -x - 3 6 4 18. y = 35 x + 6 2 19. y = 0 4 6 x –6–4 L 20. y = 2x - 4 –4 21. y = 2x 1 2 Practice 3-8 1.–3. 4.–6. Q 39. y 6 (0, 6) 4 2 –6–4–2 –2 –4 –6 40 2 4 6 x (6, 0) Answers 40. y 6 4 1 (– 2, 0) 2 (0, 2) –6–4–2 All rights reserved. 18. y 6 4 y5x 4 2 T 2 4 6 x –4 –6 Geometry Chapter 3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 17. (continued) Chapter 3 Answers (continued) 14. Sample: 7.–9. K b c 10. Sample: 15. All rights reserved. b a a 11. Sample: a b Reteaching 3-1 1a.–1b. Sample: c 60 © Pearson Education, Inc.,publishing as Pearson Prentice Hall. 12. 1c.–1d. Sample: b b 120 60 60 120 b 120 60 60 120 b 2. 110 7. 70 3. 70 8. 110 4. 110 5. 110 6. 70 Reteaching 3-2 13. 1. l6m 1 3 Given If 6 lines, then corresponding s are . a m3 = m2 = 180 Substitution c 2 and 3 are supplementary. Definition of supplementary m1+ m2 = 180 Angle Addition Postulate Geometry Chapter 3 Answers 41 (continued) 2. 2 3 Given 3 1 a6b Substitution If corresponding s, then lines are parallel. 2 1 Vertical s are . 15. y = 14 x + 21 17. y = -x + 1 Reteaching 3-7 1a. -2 1b. 12 2a. 14 3b. 0 4a. y = -2x + 4 4c. 2b. -4 3a. undefined 4b. y = 12 x +32 y Reteaching 3-3 4 1. Draw a temporary line through A that is perpendicular to , then draw a second, permanent line through A that is perpendicular to the first. This line will be parallel to . Repeat for B. The two permanent lines will be parallel not only to but to each other (Theorem 3-9). 2. Draw twoperpendicular lines through C, then for each one draw a line perpendicular to it that runs through D. The four lines will define a rectangle. 3. Draw a line that runs through both E and F, and then a second line at an acute or obtuse angle that runs through E. Draw a third line, parallel to the second andrunning through F, using the same procedure as in the Example and in Exercise 1. In the same way, draw a fourth line parallel to the first (at any desired distance from it). The four lines will define a parallelogram. 16. y = -34 x + 4 18. y = 1 2 2 O 2 4 x All rights reserved. Chapter 3 Answers 2 4 5b. y = 32 x- 4 5a. y = -32 x - 4 5c. y 4 Reteaching 3-4 2 1. ABD: mABD = 120, mADB = 30; CBE: 4 m3 = 60; m4 = 120 2 4 x 2 4 6a. y = 3x + 7 6c. 6b. y = -13 x - 3 y 4 2 Reteaching 3-5 1. 1 and 2 are interior angles; 3 and 4 are exterior angles. 2. m1 = 135; m2 = 90; m3 = 45; m4 = 90 3. 1 is an interior angle; 2 and 4are exterior angles; 3 is neither. 4. m1 = 60; m2 = 120; 2 O 4 2 O 2 4 x 2 4 Reteaching 3-6 Check students’ graphs. 1. y = 2x - 6 2. y = 13 x 3. y = -x - 3 5 4. y = 6 x + 2 5. y = -21 x + 1 6. y = 1 7. y = -72 x + 10 8. y = -x + 1 9. y = 25 x + 1 10. y = 1 11. y = -2x - 6 12. x = -3 13. y = -3x + 10 14. y = 3x - 10 42Answers 7. mJK = -1; mLM = -1; parallel 8. mJK = 23 ; mLM = 3 1 9. mJK = -6; mLM = -15 ; neither -2 ; perpendicular 3 10. mJK = -2 ; mLM = 54; neither 11. mJK = 2; mLM = 1 1 12. mJK = 5; mLM = 5; neither -2; perpendicular 13. mJK = 14; mLM = -14; parallel 14. mJK undefined; mLM = 0;perpendicular Geometry Chapter 3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. mCBE = 120, mCEB = 30, mBCE = 30; BDE: mBDE = 60, mDBE = 60, mBED = 60 2. DBE and ABC are acute, equiangular, and equilateral; ABD and CBE are isosceles and obtuse; ACE, ADE, CED, andCAD are right and scalene. 3. PQT: mPTQ = 45, mPQT = 90; PQR: mPQR = 90, mQPR = 45, mQRP = 45; RQS: mRQS = 90, mQSR = 45; SQT: mSQT = 90, mQST = 45, mSTQ = 45 4. PT = TS = RS = PR = 40 mm; PQ = QT = QR = QS = 28 mm 5. PQT, PQR, RQS, SQT, PRS, PTS, PRT, and RST areright and isosceles. Chapter 3 Answers (continued) ones, with precise control of the slant. 8. Figures A and B would be easy, because the lines have no more than two or three different slants. Figure C would be harder, because lots of different slants are used to achieve the perspective effect. Reteaching3-8 1. Sample: Z Enrichment 3-4 b 1. 48 X a 2. 2880 3. 4320 4. Angles have measures of 20, 70, or 90; AC 6 MD 6 LE 6 KF 6 JG; BM6 CH 6 NK; AH 6 PG; CM 6 DL 6 EK 6 JF 6 IG. Y Enrichment 3-5 All rights reserved. 1. 2 2. 5 7. Number 2.–4. Check students’ work. Enrichment ) * 3-1 ) 1. OE * is #) toAB . 2. 3, 5, 1, 4, 2, 6 or 4, 5, 1, 3, 2, 6; OE ) is # to AB ; if two angles are congruent and supplementary, then each measures 90°. 3. 1 > 2; Law of Reflection 4. 2 > 3; Alternate Interior Angles Theorem 5. 3 > 4; Law of Reflection 6. 1 > 4; Transitive Property of Congruence 3. 9 1.–8. © Pearson Education,Inc., publishing as Pearson Prentice Hall. y 3 1 5 A 4 T 2 D 3. 33 4. 41 Enrichment 3-3 1. If the paper shifts during drawing, lines meant to be parallel will not be parallel, and lines meant to be perpendicular will not be perpendicular. 2. The rectangular shape gives the T-square two perpendicular guideedges on which to line up. 3. Theorem 3-10, which says that if two lines are perpendicular to the same line, they are parallel to each other. A line drawn with the T-square is perpendicular to the edge on which the Tsquare is lined up. Two lines drawn with the T-square lined up on the same edge will beparallel to each other. 4. a ⊥ d. If the T-square is used to draw a line perpendicular to a vertical edge of the rectangular backing surface, and then is used to draw a line perpendicular to a horizontal edge of the backing surface, then the two lines will be perpendicular. 5. Figure A would be easy; all the linesare parallel or perpendicular. Figures B and C would be hard to draw, since the lines are not all parallel or perpendicular to each other. 6. They will all be parallel to . 7. A drafting machine can be used to draw slanted lines, not just vertical and horizontal 6 L 5. Sample: H Because mBAC = 41, mCAF = 180- 41 = 139; l 6 m because a pair of alternate interior angles are congruent. 6. CD and EF ; AK 7. Sample: CAB and IGH are corresponding angles related to parallel segments AB and GH . AK is the related transversal. 8. Sample: A, ADE, and AED form a triangle, so 180 - (43 + 76) = 61, so mAED = 61.Because AED and C are congruent corresponding angles, DE 6 BC by the Converse of the Corresponding Angles Postulate. Geometry Chapter 3 6. 27 Enrichment 3-6 Enrichment 3-2 2. 106 5. 20 3 4 5 6 7 8 9 ... n of sides Total 180 360 540 720 900 1080 1260 (n 2)180 degree measure Number of n(n 3)0 2 5 9 14 20 27 2 diagonals 8 1. x = 11 4. 14 N B M x E K Y P C S 7 R RENE DESCARTES Enrichment 3-7 1. (-3, 4) 2. (-2, 3) 3. (3, 3) 4. (-2, -2) 5. (-1, -1) 6. (4, -1) 7. (1, 0) 8. (2, -3) 9. (-1, 2) 10. (1, -4) 11. (-3, -3) 12. (2, 3) 13. (3, -2) 14. (5, 0) 15. (0, 1) 16. (4, 3) y-axis 4 A 3 N 2 G 1 L 0 1 E 2 N I 3 4 L 43 2 1 0 R A Y E T N I O P 1 2 3 4 Answers 5 x-axis 43 Chapter 3 Answers (continued) pentagons Enrichment 3-8 1., 3., and 4. hexagons A C 2. scalene, acute triangle Exercises 1, 3, and 4. All rights reserved. B Activity 3: Analyzing 5. No; refer to the answer to 6. E Activity 4: Modeling F 7. isosceles,obtuse triangle Chapter Project Activity 1: Paper Folding 5; all triangles are right isosceles; yes. Activity 2: Exploring triangle ✔ Checkpoint Quiz 1 1. Converse of Corresponding Angles Postulate 2. Alternate Interior Angle Theorem 3. Same-Side Interior Angles Theorem 4. Corresponding Angles Postulate5. Converse of Alternate Interior Angle Theorem 6. Vertical Angles Theorem 7. Converse of Corresponding Angles Postulate 8. Corresponding Angles Postulate 9. Converse of Same-Side Interior Angles Theorem 10. x = 50, y = 30, z = 65 ✔ Checkpoint Quiz 2 1. pentagon; 80 2. hexagon; x = 110; y = 983. quadrilateral; x = 94; y = 105 y y 4. 5. quadrilaterals 8 6 4 2 (2,0) 4 6 8 x 8642 2 2 4 6 8 (0,8) 44 Answers 8 6 4 (0,3) (1.2,0) 2 2 4 6 8 x 8642 2 4 6 8 Geometry Chapter 3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. D Chapter 3 Answers (continued) y 6. Chapter Test, Form B 8 6 4 242 2 4 6 8 1. true 2. true 3. false 4. true 5. false 6. m1 = 62, m2 = 62, alternate interior angles 7. m1 = 85, m2 = 95, same-side interior angles 8. m1 = 75, m2 = 75, alternate interior angles y y 9. 10. (12,0) 2 4 6 8 10 12 x (0,8) 7. Parallel; slopes are the same. 8. neither 9. Perpendicular; the product of theslopes is -1. 10. neither Chapter Test, Form A All rights reserved. 1. true 6. false 2. true 7. true 3. false 8. true 4. false 5. true 9. Answers may vary. Sample: m1 = 125, Same-Side Interior Angles Theorem; m2 = 55, Alternate Interior Angles Theorem 10. Answers may vary. Sample: m1 = 60, CorrespondingAngles Postulate then Angle Addition Postulate; m2 = 60, Same-Side Interior 11. Answers may vary. Sample: m1 = 85, Angles Theorem Alternate Interior Angles Theorem; m2 = 95, Same-Side 12. Answers may vary. Sample: Interior Angles Theorem m1 = 75, Corresponding Angles Postulate; m2 = 105,13. Answers may vary. Sample: Angle Addition Postulate m1 = 91, Corresponding Angles Postulate and Same-Side Interior Angles Theorem; m2 = 89, Corresponding Angles 14. Answers may vary. Sample: m1 = 60, Postulate Alternate Interior Angles Theorem; m2 = 115, Same-Side Interior AnglesTheorem 15. © Pearson Education, Inc., publishing as Pearson Prentice Hall. X 4 4 4 2 2 2 2 4 x 2 4 x 2 4 4 11. AD and WZ 12. AD and WZ 13. AD and WZ 14. AW and DZ 15. x = 92, y = 88 16. w = 31, x = 65, y = 65, z = 115 17. 1440 18. 45 19. neither 20. perpendicular 21. parallel 22. y = 3x - 1 23. y =14 x - 3 24. y = -2x - 2 Alternative Assessment, Form C TASK 1: Scoring Guide a. t 1 2 4 3 5 8 6 7 a b b. Sample: 8 2; 8 4; 8 and 3 are supplementary. c. m2 = 75; m3 = 105, m4 = 75, m5 = 105, m6 = 75, m7 = 105, m8 = 75 m d. 7 and 4 are supp. Given 16. x 4 8 a6b Supp. of the same are . If alt. int s are ,then lines are 6. 7 and 8 are supp. y 17. WA and XB 18. none 19. WZ and AB 20. none 21. WZ and AB 22. WZ and AB; AX and BY 23. x = 22; y = 120 24. x = 70; y = 60; z = 120 25. x = 35; y = 35; z = 55 26. 5940 27. 18 28. perpendicular 29. neither 30. parallel 31. y = 6x + 23 32. y = -21 x + 3 33. y = 13x + 2 Geometry Chapter 3 Angle Add. Post. 3 Student draws an accurate diagram and supplies correct answers and a complete and accurate flow proof. 2 Student draws a figure or gives answers that contain minor errors. 1 Student draws a figure or gives answers that contain significant errors oromissions. 0 Student makes little or no attempt. Answers 45 (continued) TASK 2: Scoring Guide TASK 4: Scoring Guide Sample: Sample: Q A P Isosceles Triangles B Equilateral Triangles 3 Student constructs an accurate figure. 2 Student constructs a figure that contains minor errors or omissions. 1Student constructs a figure that contains significant errors or omissions. 0 Student makes little or no attempt. TASK 3: Scoring Guide Sample: Triangles R 3 Student draws an accurate diagram. 2 Student draws a diagram that contains minor errors or omissions. 1 Student draws a diagram that containssignificant errors or omissions. 0 Student makes little or no attempt. All rights reserved. Chapter 3 Answers Cumulative Review B A D C 1. D 2. G 3. C 4. F 5. D 6. J 7. A 8. F 9. C 10. H 11. D 12. J 13. B 14. G 15. C 16. G 17. C 18. Given 19. Same-Side Interior Angles Theorem 20. Corresponding AnglesPostulate 21. Corresponding Angles Postulate 22. substitution 23. Check students’ work. 24. Sketches may vary. The right angle must be between the equal sides. 25. No; a triangle cannot have © Pearson Education, Inc., publishing as Pearson Prentice Hall. two sides equal and no sides equal at thesame time. a. For the figure given, y = x and y = x + 3. The lines are parallel because both lines have a slope of 1. c. For ABC, the sum of the three angles is 180°. Minor discrepancies are the result of measurement error and rounding error. d. For the figure given, ABDC is not regular. By the distanceformula, the sides are not congruent. 3 Student draws the figure accurately, writes correct equations, and reasons logically. 2 Student draws a figure, gives arguments, and writes equations that are mainly correct but may contain minor errors. 1 Student presents work with significant errors. 0 Studentmakes little or no attempt. 46 Answers Geometry Chapter 3 Chapter 4 Answers Practice 4-1 1. m1 = 110; m2 = 120 2. m3 = 90; m4 = 135 3. m5 = 140; m6 = 90; m7 = 40; m8 = 90 4. CA JS , AT SD , CT JD 5. C J, A S, T D 6. WZ JM , WX JK , XY KL , ZY ML 7. W J, X K, Y L, Z M 8. Yes; GHJ IHJ by Allrights reserved. Theorem 4-1 and by the Reflexive Property of . Therefore, GHJ IHJ by the definition of triangles. 9. No; QSR TSV because vertical angles are congruent, and QRS TVS by Theorem 4-1, but none of the sides are necessarily congruent. 10a. Given 10b. Vertical angles are . 10c. Theorem 4-

1 10d. Given 10e. Definition of triangles Practice 4-2 1. ADB CDB by SAS 2. not possible 3. not possible 4. TUS XWV by SSS 5. not possible 6. DEC GHF by SAS 7. MKL KMJ by SAS 8. PRN PRQ by SSS 9. not possible 10. C 11. AB and BC 12. A and B 13. AC 14a. Given 14b. Reflexive Property ofCongruence 14c. SAS Postulate 15. Statements Reasons 1. EF FG , DF FH 1. Given 2. DFE HFG 3. DFE HFG 2. Vertical s are . 3. SAS Postulate © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. not possible 2. ASA Postulate 3. AAS Theorem 4. AAS Theorem 5. not possible 6. notpossible 7. ASA Postulate 8. not possible 9. AAS Theorem 10. Statements Reasons 11. so UVX WVX by SSS, and U W by CPCTC. 6. ZAY and CAB are vertical angles, so ABC AYZ by ASA. ZA AC by CPCTC. 7. EG is a common side, so DEG FEG by SAS, and FG DG by CPCTC. 8. JKH and LKM arevertical angles, so HJK MLK by AAS, and JK KL by CPCTC. 9. PR is a common side, so PNR RQP by SSS, and N Q by CPCTC. 10. First, show that ACB and ECD are vertical angles. Then, show ABC EDC by ASA. Last, show A E by CPCTC. 11. First, show FH as a common side. Then, show JFH GHFby SAS. Last, show FG JH by CPCTC. Practice 4-5 1. x = 35; y = 35 2. x = 80; y = 90 3. t = 150 4. r = 45; s = 45 5. x = 55; y = 70; z = 125 6. a = 132; b = 36; c = 60 7. x = 6 8. a = 30; b = 30; c = 75 9. z = 120 10. AD ; D F 11. GA ; ACG AGC 12. KJ ; KIJ KJI 13. DC ; CDE CED 14. BA ; ABJ AJB 15. CB ;BCH BHC 16. 130 17. 65 18. 130 19. 90 20. x = 70; y = 55 21. x = 70; y = 20 22. x = 45; y = 45 Practice 4-6 Practice 4-3 1. K M, KL ML 2. JLK PLM 3. JKL PML 3. KLJ PMN by ASA, so K P by CPCTC. 4. QS is a common side, so QTS SRQ by AAS. QST SQR by CPCTC. 5. VX is a common side, 1.Given 2. Vertical s are . 3. ASA Postulate Q S Given TRS RTQ QRT STR Given AAS Theorem RT TR Reflexive Property of 12. BC EF 13. KHJ HKG or KJH HGK 14. NLM NPQ Practice 4-4 1. BD is a common side, so ADB CDB by SAS, and A C by CPCTC. 2. FH is a common side, so 1. Statements 1.AB # BC , ED # FE 2. B, E are right s. 3. AC FD , AB ED 4. ABC DEF 2. Statements 1. P, R are right s. 2. PS QR 3. SQ QS 4. PQS RSQ 3. Reasons 1. Given 2. Perpendicular lines form right s. 3. Given 4. HL Theorem Reasons 1. Given 2. Given 3. Reflexive Property of 4. HL Theorem MJN and MJK areright s. Perpendicular lines form right s. MN MK MJN MJK Given HL Theorem MJ MJ Reflexive Property of FHE HFG by ASA, and HE FG by CPCTC. Geometry Chapter 4 Answers 35 Chapter 4 Answers (continued) GI JI Reteaching 4-2 HI HI IHG IHJ Reflexive Property of HL Theorem GHI JHI GHI andJHI are right s. Given Theorem 2-5 mGHI + mJHI = 180 Practice 4-7 1. ZWX YXW; SAS 2. ABC DCB; ASA 3. EJG FKH; ASA 4. LNP LMO; SAS 5. ADF BGE; SAS 6. UVY VUX; ASA B C B C 7. common side: BC D 8. F G E G common side: FG F H 9. common angle: L L L N K J M 10. Sample:Statements 1. AX AY 2. CX # AB, BY # AC 3. mCXA and mBYA = 90 4. A A Reasons 1. Given 2. Given 3. Perpendicular lines form right s. 4. Reflexive Property of 5. BYA CXA 5. ASA Postulate 11. Sample: Because FH GE, HFG EGF, and FG GF, then FGE GFH by SAS. Thus, FE GH by CPCTC andEH HE, then GEH FHE by SSS. Reteaching 4-1 1. b 36 2. c 3. a 4. 117 Answers 5. 119 3. SSS; AEB CDB 5. SSS; PRQ VUT 7. SAS; QSP QSR Reteaching 4-3 1. 2. Angle Addition Postulate 5. RS VW 6. none 7. mC and mF = 90 8. GH JH 9. LN PR 10. ST UV or SV UT 11. mA and mX = 90 12. mF andmD = 90 13. GI # JH A 1.–2. Check students’ work. 4. SAS; MNQ ONP 6. SSS: JMK LMK 8. SAS; YTX WXT or 3. Check students’ work. 4. ABD CBD 5. JMK LKM or JKM LMK 6. UZ YZ 7. DY DO 8. P A 9. CYL ALY All rights reserved. Given Reteaching 4-4) 1a. QK QA; QB bisects KQA 1b. definition 1c.BQ BQ 1d. SAS Postulate of bisector 1e. CPCTC 2. Statements Reasons 1. MN MP, 1. Given NO PO 2. MO MO 2. Reflexive Property of 3. MPO MNO 3. SSS Postulate 4. N P 4. CPCTC 3. Statements Reasons 1. ON bisects JOH, 1. Given J H 2. JON HON 2. Definition of bisector 3. ON ON 3. ReflexiveProperty of 4. JON HON 4. AAS Theorem 5. JN HN 5. CPCTC Reteaching 4-5 1. Each angle is 60°. 2. 120 3. 120 4. 50 5. 70 6. 60 7. 65 8. 115 9. 55 10. 120 11. 60 Reteaching 4-6 1. Sample: RS = 1.3 cm, ST = 1.6 cm, QT = 2.5 cm, QR = 2.3 cm; not congruent 2. Sample: NT = 2.3 cm, TG = 2.3 cm, AT =1.9 cm; NAT GAT 3. Sample: TO = 3.3 cm, TR = 2.8 cm, MO = 2.8 cm; TOM OTR 4. HL Theorem can be applied; BDA CAD. 5. HL Theorem cannot be applied. 6. HL Theorem can be applied; MUN MLN. 7. HL Theorem can be applied; THF FET or THF TEF. 8. HL Theorem can be applied; OKR AHR. 9.HL Theorem cannot be applied. Geometry Chapter 4 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4. Chapter 4 Answers (continued) Reteaching 4-7 All rights reserved. 1. Statements 1. PSR and PQR are right s; QPR and SRP 2. PSR and PQR 2. Right s are congruent. 3. PR PR 3.Reflexive Property of 4. QPR SRP 4. AAS Theorem 5. STR QTP 5. Vertical s are . 6. PQ RS 6. CPCTC 7. STR QTP 7. AAS Theorem 2. Sample: Prove MLP QPL by the AAS Theorem. Then use CPCTC and vertical angles to show MLN QPN by the AAS Theorem. 3. Sample: Prove ACD ECB by the SASPostulate. Then use CPCTC and vertical angles to show ABF EDF by the AAS Theorem. Enrichment 4-1 Check students’ work. Samples shown. 1. 2. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Enrichment 4-2 Reasons 1. Given 3. 4. 1a. Definition of perpendicular lines 1b. AKF GEL1c. SAS 2a. Segment Addition Postulate 2b. LR + RG = TF + TA 2c. RG TA 2d. Alternate Interior Angles 2e. Corresponding Angles 2f. DAT JGR 2g. SAS Enrichment 4-3 1.–11. Check students’ work. 2a. ASA 2b. The top angles are congruent because the fold bisected the right angles formed by the foldsin steps 1 and 3. The corners of the paper are right angles; therefore, those angles are congruent. The included sides are congruent because the fold in step 1 found the midpoint of the width of the paper, thus creating two equal segments. 3a. ASA 3b. The top angles are congruent because the foldbisected the right angles formed by the folds in steps 1 and 2. The upper corners that became inside angles along the center line are right angles; therefore, those angles are congruent. The included sides are congruent because the fold in step 1 found the midpoint of the width of the paper, thus creatingtwo equal segments. 4a. The top angles are congruent because the fold bisected the right angles formed by the fold in step 1. The inside angles along the center line are right angles because the horizontal fold that formed them is perpendicular to the original fold in step 1. 4b. The included sides arecongruent because the fold in step 1 found the midpoint of the width of the paper, thus creating two equal segments. 8a. ASA 8b. The top angles are congruent because the fold bisected the right angles formed by the fold in step 7. The inside angles along the center line are congruent because of the AngleAddition Postulate. The included sides are congruent because the fold in step 7 found the midpoint of the width of the paper, thus creating two equal segments. Enrichment 4-4 1. ABT 2. ACT 3. 45; 45; ABT; ACT 4. 30; 30; ATB; ATC 5. Reflexive Property of 6. AAS Theorem 7. CPCTC 8. Definition ofsegments 9. Definition of segments 10. Definition of segments 11. SSS Postulate 12. CPCTC 13. 60 Enrichment 4-5 1. 60 2. 60 3. 60 4. 70 5. 70 6. 40 7. 72 8. 72 9. 36 10. 30 11. 30 12. 120 13. 80 14. 80 15. 20 16. 80 17. 80 18. 20 19. 41 20. 23 21. 116 22. 23 23. 41 24. 116 25. 80 26. 80 27. 20 28. 8229. 82 30. 16 31. 78.5 32. 78.5 33. 37.5 34. 40 35. 40 36. 100 5. Geometry Chapter 4 Answers 37 Chapter 4 Answers (continued) Enrichment 4-6 Chapter Project 1 Activity 1: Modeling I H E G 5H O S Y N O P T Activity 2: Observing G S O E Check students’ work. 9A R C U 12 V E U E R T I C A 14 A AB 16 A C E S 15 R E A C O T N E F L E E X I Activity 3: Investigating 11 N N N M P U M S L R 10 E E T 7H 13 L N T 6A T tetrahedron Sample: You could add in diagonals of the cube. Check students’ work. V V E N R T 17 E E S A Finishing the Project S R I O R Enrichment 4-7 1. Sample: ABD AEC 2.A A E D 5. 8 7. K H J L N J H M common side: IM M 11. 4 12. PUV TWV by AAS; PSV TQV by AAS; PVW TVU by SSS; WQV USV by AAS. 13. 14 14. Check students’ work. 15. Check students’ work. 10. 4 38 Answers 1. ABC, ABD 2. Hypotenuse-Leg Theorem 3. CPCTC 4. S Q, RQ TS, STR QRT 5a.definition of a bisector 5b. Reflexive 5c. ASA 5d. CPCTC 5e. definition 1. x = 50; y = 65 2. a = 118; b = 62; c = 59 3. HL 4. not possible 5. SAS 6. AAS 7. ASA 8. SSS 9. not possible 10. SSS 11. not possible 12. Check students’ work; J P, 6. Sample: GNJ KLH 8. Sample: HIM JIM I I 9. LN TR, L T, M Q, NR 8. Alternate Interior Angles Theorem 9. Alternate Interior Angles Theorem 10. ASA Chapter Test, Form A C G 2. SAS 3. SSS 4. not possible 6. not possible 7. LM TQ, MN QR, common side: DC C D 1. AAS 5. AAS ✔ Checkpoint Quiz 2 C 3. Sample: DEC CBD 4. B ✔ Checkpoint Quiz 1 common angle:A B E D Check students’ work. All rights reserved. 8L triangle Yes; the brace makes two rigid triangles. 3R 4 I K Q,L R, JK PQ , KL QR , JL PR . 13. B 14. F 15. C 16a. Given 16b. Given 16c. Converse of Isosceles Triangle Theorem 16d. SAS Postulate 16e. CPCTC 16f. Isosceles Triangle Theorem 17.Sample: Statements Reasons 1. BD # AC ; D is 1. Given midpoint of AC 2. BDC BDA 2. Perpendicular lines form right s. 3. AD CD 3. Definition of midpoint 4. BD BD 4. Reflexive Property of 5. BAD BCD 5. SAS Postulate 6. BC BA 6. CPCTC 18. Sample: Given that X is the midpoint of AD and BC, AX DXand BX CX by the definition of midpoint. AXB DXC because all vertical angles are congruent. AXB DXC by the SAS Postulate, and therefore AB DC by CPCTC. Geometry Chapter 4 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2C Chapter 4 Answers (continued) Chapter Test, Form B1. x = 58, y = 64 2. a = 40, b = 70, c = 70 3. SSS 4. SAS 5. HL 6. AAS 7. not possible 8. not possible 9. SSS or SAS 10. SSS or SAS 11. ASA 12. A D TASK 2: Scoring Guide a. Sample: L 5 cm 4 cm O K 4 cm M 5 cm N C B F E b. Sample: Using the Pythagorean theorem, show that All rights reserved.kABC O kDEF A D; B E; C F; AB DE; AC DF; BC EF 13. B 14. AB XY; BC YZ; AC XZ 15. A X; B Y; C Z 16. C 17. J 18a. Given 18b. Given 18c. Def. of equilateral 18d. Reflexive Property of 18e. SSS Postulate Alternative Assessment, Form C N kMNO O kOPM P O A © Pearson Education, Inc., publishingas Pearson Prentice Hall. ASA: kABE O kACD D E B C P SAS: kPOS O kPOR S O Sample: Statements 1. AE AD, B C 2. A A 3. ABD ACE 4. AB AC 5. EB DC Reasons 1. Given 2. Reflexive Property of 3. AAS Theorem 4. CPCTC 5. Segment Addition Postulate 3 Student gives a proof that is accurateand complete. 2 Student gives a proof that contains minor errors or omissions. 1 Student gives a proof that contains significant errors or omissions. 0 Student makes little or no attempt. TASK 4: Scoring Guide Sample: The SSS, ASA, and SAS Postulates are statements that are accepted as true withoutproof. The HL and AAS Theorems, on the other hand, can be proved true, using postulates, definitions, and previously proved theorems. R 3 Student’s figures and information are clear and accurate. 2 Student’s figures and information contain minor errors or omissions. 1 Student’s figures and informationcontain significant errors or omissions. 0 Student makes little or no attempt. Geometry Chapter 4 3 Student’s figures and explanation are accurate and clear. 2 Student’s figures and explanation contain minor errors or omissions. 1 Student’s figures and explanation contain significant errors or omissions. 0Student makes little or no attempt. TASK 3: Scoring Guide TASK 1: Scoring Guide Sample: SSS: M KO = MO. Then KOL MON by SAS Postulate or SSS Postulate. 3 Student gives an explanation that is thorough and correct. 2 Student gives an explanation that is partially correct. 1 Student gives anexplanation that lacks demonstrated understanding of the difference between a theorem and a postulate. 0 Student makes little or no attempt. Answers 39 Chapter 4 Answers (continued) Cumulative Review 1. D 7. A 13. 2. G 8. G 3. D 9. C C 4. H 10. G 5. D 6. G 11. A 12. J D 14. x = 102; y = 102 16. AECD 15. c, e, a, b, d or e, c, a, b, d Given AED CDE SAS Theorem CED ADE CPCTC All rights reserved. AED CDE Given ED DE Reflexive Property © Pearson Education, Inc., publishing as Pearson Prentice Hall. 17. Sample: The alarm sounds if and only if there is smoke. If the alarm sounds, then thereis smoke. If there is smoke, then the alarm sounds. 18. AAA, SSA 40 Answers Geometry Chapter 4 Chapter 5 Answers Practice 5-1 Practice 5-5 1a. 8 cm 1b. 16 cm 1c. 14 cm 2a. 22.5 in. 2b. 15.5 in. 2c. 15.5 in. 3a. 9.5 cm 3b. 17.5 cm 3c. 14.5 cm 4. 17 5. 20.5 6. 7 7. 128 19 8. 42 9. 16.5 10a. 18 10b. 6111. GH 6 AC, HI 6 BA, GI 6 BC 12. PR 6 YZ, PQ 6 XZ, XY 6 RQ 1. M, N 2. C, D 3. S, Q 4. R, P 5. A, T 6. S, A 7. yes; 4 + 7 8, 7 + 8 4, 8 + 4 7 8. no; 6 + 10 17 9. yes; 4 + 4 4 10. yes; 1 + 9 9, 9 + 9 1, 9 + 1 9 11. yes; 11 + 12 13, 12 + 13 11, 13 + 11 12 12. no; 18 + 20 40 13. no; 1.2 + 2.6 4.9 14. no; 8 21 +9 14 18 15. no; 2. 5 + 3.5 6 16. BC, AB, AC 17. BO, BL, LO 18. RS, ST, RT 19. D, S, A 20. N, S, J 21. R, O, P 22. 3 x 11 23. 8 x 26 24. 0 x 10 25. 9 x 31 26. 2 x 14 27. 13 x 61 All rights reserved. Practice 5-2 1. WY is the perpendicular bisector of XZ. 2. 4 3. 7.5 4. 9 5. right triangle 6. 5 7. 17 8. 17 9.equidistant 10. isosceles triangle 11. 3.5 ) 12. 21 13. 21 14. right triangle 15. JP is the bisector of LJN. 16. 9 17. 45) 18. 45 19. 14 20. Sample: Point M lies on JP . 21. right isosceles triangle Practice 5-3 1. (-2, 2) 2. (4, 0) 3. (2, 1) 4. Check students’ work. The final result of the construction is shown. ©Pearson Education, Inc., publishing as Pearson Prentice Hall. Triangles will vary. 1a. PQ = 8 1b. MN = 16 1c. YZ = 32 2a. NO = 2.5 2b. ST = 10 2c. UV = 20 3a. QR = 4 3b. ST = 8 3c. UV = 16 Reteaching 5-2 B A Reteaching 5-1 1. no 2. yes 3. yes 4. no 5. B 6. C Reteaching 5-3 C 1. Sample: 5. altitude 6.median 7. none of these 8. perpendicular bisector 9. angle bisector 10. altitude 11a. (2, 0) 11b. (-2, -2) 12a. ( 0, 0) 12b. (3, -4) 13a. (0, 0) 13b. (0, 3) Practice 5-4 1. I and III 2. I and II 3. The angle measure is not 65. 4. Tina does not have her driver’s license. 5. The figure does not have eight sides. 6. Therestaurant is open on Sunday. 7. ABC is congruent to XYZ. 8. mY 50 9a. If two triangles are not congruent, then their corresponding angles are not congruent; false. 9b. If corresponding angles are not congruent, then the triangles are not congruent; true. 10a. If you do not live in Toronto, then you do notlive in Canada; false. 10b. If you do not live in Canada, then you do not live in Toronto; true. 11. Assume that mA 2 mB. 12. Assume that TUVW is not a trapezoid. 13. Assume that LM does not intersect NO. 14. Assume that FGH is not equilateral. 15. Assume that it is not sunny outside. 16. Assume that Dis obtuse. 17. Assume that mA 90. This means that mA + mC 180. This, in turn, means that the sum of the angles of ABC exceeds 180, which contradicts the Triangle Angle-Sum Theorem. So the assumption that mA 90 must be incorrect. Therefore, mA 90. Geometry Chapter 5 2. Sample: Reteaching 5-41. Step 1: B; Step 2: D; Step 3: F; Step 4: E; Step 5: A; Step 6: C 2. Step 1: Assume EF DE. Step 2: If EF DE, then by the Isosceles Triangle Theorem, D F. Step 3: But D F. Step 4: Therefore, EF DE. Reteaching 5-5 1. Check students’ work. The longest side will be opposite the largest angle. The shortestside will be opposite the smallest angle. 2. largest: DEF; smallest: DFE 3. largest: PQR; smallest: PRQ 4. largest: ACB; smallest: CBA 5. longest: DF; shortest: FE 6. longest: PQ; shortest: RQ 7. longest: SV; shortest: ST Answers 29 Chapter 5 Answers (continued) 3. centroid 4. interior 5. It is not possiblefor the centroid to be on the exterior because all the medians are in the interior of the triangle. Enrichment 5-1 1., 5. y A 6. 4 E F D 2 H F 4 C6 x 2E B 2 2. D(0, 3) 3. E(2.5, 0) 4. F(2.5, 3) 6. 3.75 square units 7. They are equal. yQ 9., 13. 8. 1 : 4 2 All rights reserved. 4 V T 2 4 6 8 10 12 x I 2 R 4 U S 7.orthocenter 10. T(3.5, 0.5) 11. U(9, -3) 12. V(6.5, 0.5) 14. 21 square units 15. 5.25 square units 8. It is possible for the orthocenter to be in the interior, the exterior, or on the triangle itself. Explanations may vary. 9. X Center of Gravity Enrichment 5-2 1.–10. Check students’ work. Y 10. Check students’work. 11. centroid Enrichment 5-4 11. Check students’ work. Give hint: mAOB = 45°. A O 1. If a fish is a salmon, then it swims upstream. 2. If a fish swims upstream, then it is a salmon. 3. If a fish is not a salmon, then it does not swim upstream. 4. If a fish does not swim upstream, then it is not a salmon.5. Exercises 1 and 4 6. the salmon 7. the fish that swim upstream 8. 9. U B Q P P 10. Enrichment 5-3 1.–2. A U Q U 11. U Q Q P P D B 30 C Answers Geometry Chapter 5 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Z Chapter 5 Answers (continued) 12. If a quadrilateral is a rectangle,then the angles of the quadrilateral are congruent. 13. If the angles of a quadrilateral are not congruent, then the quadrilateral is not a rectangle. 14. If a quadrilateral is not a rectangle, then the angles of the quadrilateral are not congruent. 15. All are true (12, 13, and 14). 16. the contrapositive Enrichment5-5 1. U 8. T 14. F 2. T 3. F 4. T 5. U 6. F 7. T 9. T 10. F 11. T 12. U 13. T 15. F 16. U 17. T 18. T The name of the mathematician is THALES (th –a l –ez). All rights reserved. Chapter Project Activity 1: Creating isoceles; median, angle bisector, perpendicular bisector 1. 4 2. 4.5 3. 6.5 4. 5 5. Sample: VP =RT 6. D, B, C 7. C, D, B 8. C 9. 3; 21 10. inside 11. inside 12. on 13. BC 14. LO, LM, MO, MN, NO 15. 3 16. 25 17. 6 Check students’ work. Activity 3: Designing Check students’ work. ✔ Checkpoint Quiz 1 1. angle ) bisector, since TM NM © Pearson Education, Inc., publishing as Pearson Prentice Hall.angle. This means that one angle measures 90. This, in turn, means that each of the three angles is 90 because the triangle is equilateral and also equiangular. This means that the angle sum of the triangle is 270, contradicting the Triangle AngleSum Theorem. So the assumption that an equilateraltriangle can have a right angle must be incorrect. Therefore, an equilateral triangle cannot have a right angle. 14. Each leg must be greater than 5 units in length. 15. 6; 30 16. on 17. inside 18. outside 19. AB 20. XD, XC, XB 21. (5, 4) 22. (1, -2) 23. 2 24. 15 25. 24 Chapter Test, Form B Activity 2:Experimenting QM is the angle bisector 4. 6 5. 4 6a. 18 8. congruent 3a. If two lines are not parallel, then they intersect; false. 3b. If two lines intersect, then they are not parallel; true. 4. x = 9.5 5. x = 5 6. Sample: EF HG 7. I and III 8. II and III 9. D, B, C 10. B, D, C 11. C, D, B 12. C 13. Sample: Assumethat an equilateral triangle has a right 2. congruent, since 3. 20, since TM NM 6b. 42 7. altitude ✔ Checkpoint Quiz 2 1. Inv: If you don’t eat bananas, then you will not be healthy. Contra.: If you are not healthy, then you don’t eat bananas. 2. Inv.: If a figure is not a triangle, then it does not have three sides.Contra.: If a figure does not have three sides, then it is not a triangle. 3. Suppose that there are two obtuse angles in a triangle. This would mean that the sum of two angles of a triangle would exceed 180°. This contradicts the Triangle-Sum Theorem. Therefore, there can be only one obtuse angle in atriangle. 4. Suppose that the temperature is below 32°F and it is raining. Because 32°F is the freezing point of water, any precipitation will be in the form of sleet, snow, or freezing rain. This contradicts the original statement. Therefore, the temperature must be above 32°F for it to rain. 5. yes; 6 + 4 8, 8 + 64 6. no; 2.6 + 4.1 6.7 7. B, C, A 8. D, F, E Alternative Assessment, Form C TASK 1: Scoring Guide Sample: Assume that Harold can create a triangular garden plot with sides of 6 ft, 6 ft, and 13 ft. Then a triangle is possible with sides 6, 6, and 13. But 6 + 6 13, so by the Triangle Inequality Theorem, nosuch triangle exists. Therefore, the plan is impossible. 3 Student gives a clear and accurate explanation. 2 Student gives an explanation that contains minor inaccuracies. 1 Student gives an explanation that contains significant flaws in logical reasoning. 0 Student makes little or no attempt. TASK 2:Scoring Guide Sample: The locus of points in a plane equidistant from a single point forms a circle, as shown in the first figure. The locus of points in a plane equidistant from the endpoints of a segment forms a perpendicular bisector of the segment, as shown in the second figure. Chapter Test, Form A1a. If a triangle does not have three congruent sides, then it is not equiangular; true. 1b. If a triangle is not equiangular, then it does not have three congruent sides; true. 2a. If an isosceles triangle is not obtuse, then the vertex angle is not obtuse; true. 2b. If the vertex angle of an isosceles triangle is notobtuse, then the triangle is not obtuse; true. Geometry Chapter 5 Answers 31 (continued) 3 Student gives an accurate explanation and drawings. 2 Student gives a drawing or an explanation that contains minor errors. 1 Student gives an explanation or a drawing that contains significant errors. 0 Studentmakes little or no attempt. TASK 3: Scoring Guide Expressed as a conditional: If ABC is an equilateral triangle with midpoints D, E, and F on sides AB, BC, and AC respectively, then DEF forms an equilateral triangle. Sample: DEF is an equilateral triangle. Each of DEF’s sides is exactly 12 thecorresponding side of ABC. If ABC’s sides are all the same length, then all of DEF’s sides are all the same length as well. Cumulative Review 1. A 8. H 14. F 2. G 3. D 4. H 5. B 6. G 7. D 9. D 10. F 11. A 12. J 13. B 15. 24 16. 28 17. Sample: The principal is giving a student an award. 18. Suppose that anequilateral triangle has an obtuse angle. Then one angle in the triangle has a measure greater than 90. But this contradicts the fact that in an equilateral triangle, the measure of each angle is 60. Therefore, the equilateral triangle does not have an obtuse angle. 19. Sample: 8 8 6 4 8 10 3 Studentaccurately states the conditional and the proof. 2 Student has minor errors in the statement of the conditional or in the proof. 1 Student gives a statement of conditional or a proof that contains significant errors. 0 Student makes little or no attempt. 20. (-1, 2) 21. 4 All rights reserved. Chapter 5 Answers T QM TASK 4: Scoring Guide Sample: 22. C, B, A B A © Pearson Education, Inc., publishing as Pearson Prentice Hall. D C ADC is isosceles. BAC = BCA because ABC is an isosceles triangle with AB = BC. The angle bisectors create two new angles at each vertex of ABC, each of 1 which has a measure 2that of the angle bisected. Thus 1 mDAC = 2 mBAC and mDCA = 12 mBCA. Because mBAC = mBCA, mDAC = mDCA, which makes ADC an isosceles triangle. 3 2 1 0 Student gives accurate answers and justification. Student gives answers containing minor errors. Student gives a justification containingmajor logical gaps. Student makes little or no effort. 32 Answers Geometry Chapter 5 Chapter 6 Answers Practice 6-1 Practice 6-5 1. parallelogram 2. rectangle 3. quadrilateral 4. parallelogram, quadrilateral 5. kite, quadrilateral 6. rectangle, parallelogram, quadrilateral 7. trapezoid, isosceles trapezoid,quadrilateral 8. square, rectangle, parallelogram, rhombus, quadrilateral 9. x = 7; AB = BD = DC = CA = 11 10. m = 9; s = 42; ON = LM = 26; OL = MN = 43 11. f = 5; g = 11; FG = GH = HI = IF = 17 12. parallelogram 13. rectangle 14. kite 15. parallelogram 1. 118; 62 5. 101; 79 © Pearson Education, Inc.,publishing as Pearson Prentice Hall. All rights reserved. Practice 6-2 1. 15 2. 32 3. 7 4. 8 5. 12 6. 9 7. 8 8. 358 9. 54 10. 34 11. 54 12. 34 13. 100 14. 40; 140; 40 15. 70; 110; 70 16. 113; 45; 22 17. 115; 15; 50 18. 55; 105; 55 19. 61; 72; 108; 32 20. 32; 98; 50 21. 16 22. 35 23. 28 24. 4 2. 99; 81 3. 59; 1214. 96; 84 6. 67; 113 7. x = 4 8. x = 16; y = 116 9. x = 1 10. 105.5; 105.5 11. 90; 25 12. 118; 118 13. 90; 63; 63 14. 107; 107 15. 90; 51; 39 16. x = 8 17. x = 7 18. x = 28; y = 32 Practice 6-6 1. (1.5a, 2b); a 2. (1.5a, b); " a2 1 4b2 3. (0.5a, 0); a 2 4. (0.5a, b); " a 1 4b2 5. 0 6. 1 7. -21 8. 2 9. 2b 10. -2b 11. 2b12. -2b 3a 3a 3a 3a 13. E(a, 3b); I(4a, 0) 14. O(3a, 2b); M(3a, -2b); E(-3a, -2b) 15. D(4a, b); I(3a, 0) 16. T(0, 2b); A(a, 4b); L(2a, 2b) 17. (-4a, b) 18. (-b, 0) Practice 6-7 p q p = qx Practice 6-3 1a. 1. no 2. yes 3. yes 4. no 5. yes 6. yes 7. x = 2; y = 3 8. x = 6; y = 3 9. x = 64; y = 10 10. x = 8; the figure is abecause both pairs of opposite sides are congruent. 11. x = 40; the figure is y not a because one pair of opposite angles is not congruent. 12. x = 25; the figure is a because the congruent opposite sides are 6 by the Converse of the Alternate Interior Angles Theorem. 13. Yes; the diagonals bisect eachother. 14. No; the congruent opposite sides do not have to be 6. 15. No; the figure could be a trapezoid. 16. Yes; both pairs of opposite sides are congruent. 17. Yes; both pairs of opposite sides are 6 by the converse of the Alternate Interior Angles Theorem. 18. No; only one pair of opposite angles iscongruent. 19. Yes; one pair of opposite sides is both congruent and 6. 20. No; only one pair of opposite sides is congruent. Practice 6-4 1a. rhombus 1b. 72; 54; 54; 72 2a. rectangle 2b. 72; 36; 18; 144 3a. rectangle 3b. 37; 53; 106; 74 4a. rhombus 4b. 59; 90; 90; 59 5a. rectangle 5b. 60; 30; 60; 30 6a.rhombus 6b. 22; 68; 68; 90 7. Yes; the parallelogram is a rhombus. 8. Possible; opposite angles are congruent in a parallelogram. 9. Impossible; if the diagonals are perpendicular, then the parallelogram should be a rhombus, but the sides are not of 10. x = 7; HJ = 7; IK = 7 equal length. 11. x = 7; HJ = 26;IK = 26 12. x = 6; HJ = 25; IK = 25 13. x = -3; HJ = 13; IK = 13 14a. 90; 90; 29; 29 14b. 288 cm2 15a. 70; 90; 70; 70 15b. 88 in.2 16a. 38; 90; 90; 38 16b. 260 m2 17. possible 18. Impossible; because opposite angles are congruent and supplementary, for the figure to be a parallelogram they must measure90, the figure therefore must be a rectangle. Geometry Chapter 6 p 1b. y = mx + b; q = q (p) + b; b + q p2 p2 p 1c. x = r + p +q- q p2 rp p2 q; 1d. y = q (r + p) p2 rp + q - q ; y = q + q + q - q ; y = q + q; rp 1e. intersection at (r + p, q + q) 1g. y = mx + b; q = r q (r) + r q b; b = q - 1f. (r, q) r2 q; 2 y = qx + q - q1h. y = qr (r + p) + q - rq ; 2 2 rp rp y = rq + q + q - rq ; y = q + q; intersection at rp rp (r + p, q + q) 1i. (r + p, q + q) 2a. (-2a, 0) b 3a b 2b. (-a, b) 2c. (- 2 , 2 ) 2d. a 3a. (-4a, 0) 3b. (-2a, 3a) 3c. 32 3d. (2a, -a) 1 3e. 2 4. The coordinates for D are (0, 2b). The r r2 coordinates for C are (2a, 0). Given thesecoordinates, the lengths of DC and HP can be determined: DC = " (2a 2 0) 2 1 (0 2 2b) 2 = " 4a2 1 4b2; HP = " (0 2 2a) 2 1 (0 2 2b) 2 = " 4a2 1 4b2; DC = HP, so DC HP. Reteaching 6-1 1.–4. Samples: 1. trapezoid 2. parallelogram 3. rectangle Answers 35 Chapter 6 Answers (continued) Reteaching 6-3 5.parallelogram 6. rectangle 7. isosceles trapezoid 8. rhombus 9. trapezoid 10. kite 11. rectangle 12. square 3. no 4. yes 1. BD CD, AE BD, AE 6 CD 2. AE CD 3. ACDE is a parallelogram. Reteaching 6-2 1. Statements 1. Parallelogram ABCD 2. AB CD, BC DA 3. BD DB 4. ABD CDB 5. A C 2. Statements 1.Parallelogram ACDE; CD BD 2. C E 3. CBD C 4. CBD E 3. Statements 1. Parallelogram ACDE; AE BD 2. AE CD 3. CD BD 4. CBD C 4. Statements 1. Parallelogram ACDE; CBD E 2. E C 3. CBD C 4. CD BD 5. BDC is isosceles. 5. Statements 1. Isosceles trap. ABDE; C E 2. AE BD; E BDE 3. C BDE 4.CBD BDE 5. C CBD 6. BCD is isosceles. 7. BD CD 8. AE CD 36 Answers Reasons 1. Given 2. Opposite sides of a parallelogram are congruent. 3. Reflexive Prop. of 4. SSS 5. CPCTC Reasons 1. Given 2. Opposite angles of a parallelogram are . 3. Isosceles Triangle Theorem 4. Substitution Reasons 1.Given 2. Opposite sides of a parallelogram are . 3. Substitution 4. Isosceles Triangle Theorem Reasons 1. Given 2. Opposite angles of a parallelogram are . 3. Substitution 4. If 2 s of a are , sides opposite them are . 5. Def. of isosceles triangle Reasons 1. Given 2. Definition of isosceles trapezoid 3.Transitive Property 4. Alt. int. ’s are . 5. Transitive Property 6. Definition of isosceles triangle 7. Definition of isosceles triangle 8. Transitive Property 6. Statements 1. CBD C, AE BD, AC ED 2. BD CD 3. AE CD 4. ACDE is a parallelogram. Reasons 1. Given 2. Substitution 3. If one pair of opposite sides isboth congruent and parallel, then the quadrilateral is a parallelogram. Reasons 1. Given 2. If 2 s of a are , sides opposite them are . 3. Substitution 4. If both pairs of opposite sides are , then the quad. is a parallelogram. All rights reserved. 1. no 2. no 5. Statements rhombus Reteaching 6-4 1. m1 = 60; m2= 30; m3 = 90 m2 m2 m2 m3 = = = = 50; m3 = 50; m4 = 100 100; m3 = 40; m4 = 40 60; m3 = 60 5. m1 = 15; m4 = 90 6. m1 = 2. m1 = 80; 3. m1 = 80; 4. m1 = 60; 75; m2 = 75; 45; m2 = 45 Reteaching 6-5 1. 90 2. 52 7. 27 8. 63 10. Sample: 3. 90 9. 27 4. 38 5. 52 6. 90 Statements Reasons 1. LP MN 1.Given 2. LPN MNP 2. Theorem 6-15 3. PN PN 3. Reflexive Prop. of 4. LNP MPN 4. SAS Postulate 5. PM LN 5. CPCTC 6. LM LM 6. Reflexive Prop. of 7. PLM NML 7. SSS Postulate 8. LPQ MNQ 8. CPCTC 9. LQP MQN 9. Vertical angles are . 10. LQP MQN 10. AAS Theorem 11. 87 12. 48 13. 45 14. 4815. 24 16. 24 17. 132 18. 24 19. 132 20. Sample: Both LMQ and PNQ have the same angle measures, but their sides have different lengths. Reteaching 6-6 1. Q(x + k, m) 2. X(-a, 0); W(0, -b) 3. S(a, -a); T(0, -a) 4. Each side has length a" 2, so it is a rhombus. One pair of opposite sides has slope of 1, andthe other pair has slope of -1. Therefore, because (1)(-1) = -1, the rhombus has four right angles and is a square. 5. Each side Geometry Chapter 6 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4. (continued) has length of " 2a 2 1 2a 1 1. Therefore, the figure is a rhombus. 6. C(x - k, m)Reteaching 6-7 All rights reserved. 1. Each diagonal has length " c2 1 (b 1 a) 2. 2. The a c midpoints are (b2, 2c ) and (b 1 , ). The line connecting the 2 2 midpoints has slope of 0 and is therefore parallel to the third side. 3. The midpoints are (a2, 0), (a, b2), (a2, b) and (0, b2). The segments joining themidpoints each have length 1 2 2 4. The midpoints are (a2, b2), (-a2, b2), 2 "a 1 b . a b a (-2, -2 ), and (2, -b2). The quadrilateral formed by these points has sides with slopes of 0, 0, undefined, and undefined. Therefore, the sides are vertical and horizontal, and consecutive sides are perpendicular. 5. Themedian meets the base at (0, 0), the midpoint of the base. Therefore, the median has undefined slope; i.e., it is vertical. Because the base is a horizontal segment, the median is perpendicular to the base. 6. The midpoints are (a2, 0), (a 12 d, 2e ), (b 12 d, c 12 e), and (b2, 2c ). One pair of opposite sideshas slope of de , c and the other pair of opposite sides has slope of b 2 a. Therefore, the figure is a parallelogram because opposite sides are parallel. 1. Some 2. No 3. Some 6. Some 7. No 8. Some 11. Some 12. No 13. No 16. All 17. Some 18. No 4. All 5. No 9. All 10. Some 14. Some 15. All 19. Some20. All Enrichment 6-2 1. ABED, BCFE, DEHG, EFIH, ACIG, DBFH 2. ABED, BCFE, DEHG, EFIH, ACIG, DBFH, ACFD, DFIG, ABHG, BCIH 3. ABED, BCFE, DEHG, EFIH, ACIG, DBFH, ACFD, DFIG, ABHG, BCIH, AEHD, DBEG, BFIE, ECFH, BFEA, BCED, EFHG, DEIH 4. DBHG, ECIH, ADEC, EACF,ABHD, BCIE, HDFI, GDFH, BCGD, GCFH, DAIH, ABFI 5. pentagon, scalene triangle, two rectangles, two trapezoids, two isosceles right triangles 6. B 74° 62° A C 44° 44° 44° 92° D 62° 62° 56 44° F 62° E Because AJ = AB - BJ, AJ = 4.64 - 2.4 = 2.24 cm. 2. 44; LMNO is a rhombus, so MO # LN. SomMRL = 90 and mMLR = 46; therefore mMLR = 44. 3. 2.24 cm; because mAML = 46 = mMLR, AJ 6 LN. So AJLN is a parallelogram, and AJ = LN = 2.24 cm. 4. 2.4 cm; because mMLA = 44 = mLMR, AD 6 MO. AMOD therefore is a parallelogram, so MO = AD = 2.4 cm. 5. 44; because mLMR = mRMN,mRMN = 44. 6. 90; because JBCK is a square, mBJK = 90. Because mBJK + mNJM = 180, mNJM = 90. 7. 88; mLMR + mRMN = mLMN = 88. 8. No; the base angles of the triangle are not congruent. 9. parallelogram 10. rhombus 11. square 12. rectangle Enrichment 6-5 1. 64 square units 2. 65 squareunits 3. When you actually cut out the shapes and reassemble them, you find that the “diagonal” of the rectangle is not a straight line. Trapezoid IV does not quite meet the top edge of triangle I, and similarly there is a space between trapezoid III and triangle II. The extra space represents the extra onesquare unit of area. 4. The areas differ by 4 square inches. 10 in. 6 in. 10 in. 1a. Given 1b. Definition of a regular hexagon 1c. SAS 1d. CPCTC 1e. Given any two distinct 1. 2.24 cm; JBCK is a square, so BC = BJ = 2.4 cm. 6 in. Enrichment 6-3 Enrichment 6-4 10 in. © Pearson Education, Inc., publishingas Pearson Prentice Hall. Enrichment 6-1 endpoints. 1f. Definition of a diagonal 1g. Definition of a regular hexagon 1h. Reflexive Property of Congruence 1i. SSS 1j. CPCTC 1k. If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel. 1l. Definition of aregular hexagon 1m. SAS 1n. CPCTC 1o. Definition of a regular hexagon 1p. SSS 1q. CPCTC 1r. If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel. 1s. Definition of a parallelogram 2. A parallelogram can be constructed in an octagon by drawingthe diagonals as shown. Other answers are possible. 6 in. Chapter 6 Answers 10 in. 6 in. points, there is a unique line segment with these points as Geometry Chapter 6 Answers 37 Chapter 6 Answers (continued) Enrichment 6-6 F D 210 4 B E 0 360 2 O 6 4 2 330 4 G 240 11. (1; 300°) 14. (4; 315°) 17.(5; 255°) 2. 2 B 4 2 A C 6 x 6 4 2 O 2 4 D rhombus 3. 8 6 change by a constant amount. A Chapter Project B 2 C Activity 1: Doing 6 4 2 DO 2 Check students’ work. 2 4 6 x 4 Activity 2: Analyzing 1. The effective area is rectangular. 2. The effective area is rectangular. The effective area is Activity 3:Researching 6 x 6 Enrichment 6-7 larger in Figure 2 than in Figure 1 because the diagonal is longer than the face. 3. You should tie the string to a vertical stick of the kite. 4. If the faces of the kite were unchanged, one diagonal of the rhombus is longer than the diagonals of the square, so the effective areawould increase. 4 8 12. (3.5; 225°) 15. (5; 120°) 18. (1; 180°) 1. (0, 7) 2. (0, 1) 3. (5, 1) 4. (5, 7) 5. The x-coordinate of each point decreased by 4. 6. (7, 0) 7. (-3, -5) 8. (-1.5, -2.5) 9. Add 3 to the y-coordinate. 10. The y-coordinate of each point decreased by 3. 11. The y-coordinate decreased by 3. 12. (3, 4)13. M(10, 6), N(2, 6) 14. rectangle 15. Either all the x-coordinates or all the y-coordinates A trapezoid 300 270 10. (5; 60°) 13. (6; 150°) 16. (2; 90°) D C 30 A I H C 1. 60 B 150 180 90 rectangle 4. 8 6 A D 4 B 2 6 4 2 O 2 4 6 x Check students’ work. 4 ✔ Checkpoint Quiz 1 1. 70, 110, 70 2. 53, 53, 52 3. 96,84, 46 4. square 5. x = 20, y = 3 6. x = 2, y = 5 7. 30.5 ✔ Checkpoint Quiz 2 1. x = 66, y = 57 2. x = 35, y = 35 3. x = 3, y=4 4. True; they are the only quadrilaterals that possess these properties. 5. False; only two triangles at a time are congruent. 6. (n + 1, m) 7. (k, 0) 38 Answers All rights reserved. 120Chapter Test, Form A C kite 5. 5 cm 6. 3 in. 7. 4 m 8. 20 9. x = 33; y = 81 10. 10 11. 20 12. x = 30; y = 30 13. 12 14. D(a, 0); E(b, c); (a 12 b, 2c ) 15. D(-c, 0); E(0, b); (-2c , b2) 16. D(0, b); E(a, 0); (a2, b2) 17. 28; 28 18. 105; 75 19. 90; 48 20. 55; 90 21. 22; 68 22. 53; 37 23. The lengths of segments AB,BC, and AC are: AB = " j2 1 k2, BC = " k2 1 l2, AC = l + j. Thus, Geometry Chapter 6 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1.–9. Chapter 6 Answers (continued) the perimeter of ABC is l + j + " j2 1 k2 + " k2 1 l2. j -2, k2 The midpoints of segments AB, BC, and AC are: M( N( l k2, 2 ), O( l 2 j 2 ,0 ), ). The lengths of segments MN, NO, and MO are: MN = 12(l + j), NO = 12 " j2 1 k2, MO = 12 " k2 1 l2 . Thus, the perimeter of MNO is 1 2 (l + + " j2 j 1 k2 + perimeter of ABC. " k2 1 l2), which is half the 24. parallelogram 25. kite 27. parallelogram 28. square 30. isosceles trapezoid 26.rectangle 29. rhombus Chapter Test, Form B All rights reserved. 1. y 6 A 4 square 2 B 6 4 2 2 2 4 Dx 6 2. © Pearson Education, Inc., publishing as Pearson Prentice Hall. y 2 kite A 10 D x = 90 (Diagonals of a kite are #.) y = 5 (Def. of isos. trapezoid) z = 75 (Base angles of isos. trap. are .) 4 2 C 4 6 4. x 810 y 8 A trapezoid B 3 2 1 0 Student gives correct answers and reasons. Student gives mostly correct answers and reasons. Student gives mostly incorrect answers and reasons. Student makes little or no effort. TASK 4: Scoring Guide a. Q = (5 - a, 5); S = (5, 5 - a) 6 4 PR C 2 8 6 4 2 2 D 3 Student givesaccurate and complete answers and diagram. 2 Student gives answers and a diagram that are mostly accurate. 1 Student gives answers or a diagram containing significant errors. 0 Student makes little or no effort. TASK 3: Scoring Guide B 6 2 Rhombuses Squares Cx 4 6 6 2 3 Student lists allstatements accurately in part a and gives the correct answers in part b. 2 Student gives mostly correct answers but with some errors. 1 Student gives answers that fail to demonstrate understanding of the properties of parallelograms. 0 Student makes little or no effort. Rectangles B 4 4 8 Samples: a. ABCD, BC AD, AB 6 CD, BC 6 AD, ABD BDC, ACD BAC, CBD BDA, CAD BCA, BE = ED, AE = EC, ABC CDA, BCD BAD, b. C, E, F parallelogram 2 D 6 4 2 2 3. TASK 1: Scoring Guide Parallelograms C 6 y A Alternative Assessment, Form C TASK 2: Scoring Guide 4 6 15. 90; 45; 45 16. 71; 71; 38 17.parallelogram 18. rhombus 19. trapezoid 20. square 2 x 4 5. 7 in. 6. 6 cm 7. 22 m 8. x = 3.5 9. x = 19; y = 123 10. x = 6 11. 78; 102 12. 90; 61 13. 64; 128 14. 90; 63; 27 Geometry Chapter 6 0 = 55 2 2 0 = 1. Slope of QS b. Slope of 5 = 5522(5a 2 2 a) = -1. Because the product of their slopes = -1, PR #QS. 3 Student gives correct coordinates and a valid proof. 2 Student gives answers or a proof that contains minor errors. 1 Student gives incorrect coordinates in part a or a poorly constructed proof in part b. 0 Student makes little or no effort. Answers 39 Chapter 6 Answers (continued) Cumulative Review1. B 2. J 3. B 4. H 5. D 6. J 7. A 8. G 9. B 10. G 11. B 12. J 13. A 14. 10 15. 38, 50, 92 16. Proof: AB BC DC AD by the definition of a rhombus. Also, AC AC. Therefore, ABC CDA by the SSS Theorem. A D C B 17. 18. A B O N L M Q D 30 C All rights reserved. T R © Pearson Education, Inc., publishing asPearson Prentice Hall. 19. Sample: The construction of the undercarriage of a bridge; it is a combination of triangles, which are the strongest geometric polygon. 40 Answers Geometry Chapter 6 Chapter 7 Answers Practice 7-1 Practice 7-5 1. 1 : 278 2. 18 ft by 10 ft 3. 18 ft by 14 ft 4. 18 ft by 16 ft 5. 8 ftby 8 ft 6. 3 ft by 10 ft 7. true 8. false 9. true 10. true 11. false 12. true 13. true 14. false 15. false 16. 12 17. 12 18. 33 19. 21 5 20. 48 21. 88 22. 6 23. 25 24. -25 3 25. 14 : 5 26. 3 : 2 27. 12 : 7 28. 83 29. 1. BE 6. BE y=4 y = 198 5 7 13 All rights reserved. Practice 7-2 1. ABC XYZ, with similarity ratio 2 : 1 2.QMN RST, with similarity ratio 5 : 3 3. Not similar; corresponding sides are not proportional. 4. Not similar; corresponding angles are not congruent. 5. ABC KMN, with similarity ratio 4 : 7 6. Not similar; corresponding sides are not proportional. 7. I 8. O 9. J 10. NO 11. LO 12. LO 13. 3.96 ft 14. 4.8 in. 15.3.75 cm 16. 10 m 17. 32 18. 53 19. 7 12 20. 4 12 21. 53 22. 37 23. 5 Practice 7-3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. AXB RXQ because vertical angles are . A R (Given). Therefore AXB RXQ by the 3 XM PX AA Postulate. 2. Because MP LW = WA = AL = 4 , MPX LWA bythe SSS Theorem. 3. QMP AMB because vertical s are . Then, because QM 2 PM AM = BM = 1 , QMP AMB by the SAS Theorem. 4. M A (Given). Because there are 180° in a triangle, mJ = 130, and J C. So MJN ACB by the AA Postulate. 5. Because AX BX AX = BX and CX = RX, CX = RX . AXB CXRbecause vertical angles are . Therefore AXB CXR by the SAS Theorem. 6. Because AB = BC = CA and AB BC CA XY = YZ = ZX, XY = YZ = ZX . Then ABC XYZ by the SSS Theorem. 7. 15 8. 55 9. 48 7 2 4 10. 85 11. 20 12. 36 13. 33 ft 3 3 Practice 7-4 1. 16 2. 8 3. "77 4. 2 "11 5. 10"2 6. 6"5 7. h 8. y 9. x10. a 11. b 12. c 13. x = 6; y = 6"3 14. x = 8 "3; y = 4 "3 15. 92 16. x = 4 "5 ; y = "55 17. x = "3; y = "6; z = "2 18. x = 8; y = 2 "2; z = 6 "2 19. 2"15 in. Geometry Chapter 7 2. EH 3. BC 4. JD 7. 16 8. 4 9. 4 10. 3 15 12. 4 13. x = 6; y = 6 15. 2 16. 4 17. 10 5. 25 2 JG JI 11. x = 14. x = 189 5 ; 25 9; Reteaching 7-1 1. 534 apples 2. 768 gallons 3. 4627 employees 4. about 4800 picture frames 5. 90 6. 2450 7. 85 8. 560 9. 58 10. 7308 11. 12 12. 20,250 13. 2500 Reteaching 7-2 1. KLM QRS; 5 : 3 2. ABC XYZ; 10 : 7 3. not similar 4. not similar 5. LMPR QICK; 1 : 1 6. not similar 7. similar with ratio 10 : 11 8. not similar9. similar with ratio 4 : 9 Reteaching 7-3 1. ABC ZYX by AA 2. not similar 3. QEU SIO by SSS 4. ABC RST by AA 5. not similar 6. BAC XQR by SAS 7. yes; by SSS or by AA 8. No; the vertex angles may differ in measure. 9. All congruent triangles are similar with ratio 1 : 1. Similar triangles are notnecessarily congruent. Reteaching 7-4 25 60 144 13 ; y = 13 ; z = 13 = 420 3. x = 4.8; y = 29 441 2. x = 400 29 ; y = 29 ; z 8 4. x = 12; y = 16 5. x = "2; y = "3; z = "6 6. x = 2"10; 120 y = "15; z = 2"6 7. 13 1.x = Reteaching 7-5 1. x = 4.5 2. x = 38 3. x = 7.2; y = 4.8 15 25 4. x = 8 ; y = 8 5. x = 9 6. x = 23 ; y =16 24 7. BN = 5 ; CN = 5 8. BY = 32 5 2 Enrichment 7-1 1. 10 lb past year. 6. $106 2. $250; costs have not changed during the 3. $240,000 4. $168,750 5. $6785 7. 360 yd 8. 40 9. 42 10. 63 Enrichment 7-2 1.–4. Check students’ work. Answers 29 Chapter 7 Answers (continued) Enrichment 7-3 4.Opposite s of a parallelogram are . 5. UTV S 5. Opposite s of a parallelogram are . 6. QRS VUT 6. AA Similarity Post. 3. Using the distance formula, AB = "180 or 6 "5, BC = "45 or 3 "5, CA = 15, ST = 10, TB = "80 or 4 "5, and BS = "20 or 2 "5. If the two triangles are similar, then corresponding sides areproportional. Therefore, 6"5 3"5 BC CA 3 3 15 3 AB ST = 10 = 2 = TB = 4"5 = 2 = BS = 2"5 = 2. BC AB Because CA ST = TB = BS , ABC TBS by the SSS Similarity Postulate. Enrichment 7-4 1. The triangles have two congruent angles; AA Similarity (DE 2 BC)AB 1 BD = AB 3. BD = Postulate. 2. AB DEBC BC 4. 800 yd 5. 4500 ft 6. 43 ft 7. 9 m Enrichment 7-5 1. 18 2. 1 3. 19 4. 3 5. 8 6. 7 7. 22 8. 9 9. 13 10. 4 11. 5 12. 6 13. 12 14. 17 15. 2 16. 10 17. 21 18. 15 19. 20 20. 16 ARCHIMEDES OF SYRACUSE Chapter Project Activity 1: Doing Check students’ work. Activity 2: Analyzing Stage 1: 12; 4 Stage2: 19; 5 31 1 1 Stage 3: 192; 27 ; 79 Activity 3: Doing Check students’ work. 30 Answers The entire snowflake at each stage will be inside the regular hexagon. The area of the hexagon is 2 square units, so the area of the snowflake is always less than 2 square units. Activity 5: Modeling Check students’work. ✔ Checkpoint Quiz 1 6 1. 12 2. 1 : 48 3. 11 8 7. 3.6 ft 8. no; 20 2 73 4. 6 5. O 9. 780 mi 6. MO ✔ Checkpoint Quiz 2 1. yes; AA similarity postulate postulate 3. 5 4. 13.7 6. 8 13 7. 2.5 2. yes; SAS similarity 5. x = 4.5, y = 7.5 Chapter 7 Test, Form A 24 19 2 3 5. ABC XYZ by the AA Postulate 6. notsimilar 7. KPM RPQ by the SAS Theorem 8. 30 ft 9. 9 3 10. 6 "5 11. B 12. 12 13. 2 14. 6 15. 18 16. Check students’ work. 17. If two parallel 7 lines are cut by a transversal, alternate interior angles are congruent. Therefore, ABC EDC by the AA Postulate. 8 18. x = 9; y = 6 "3; z = 3 "3 19. x = 10 3;y = 3 20.C 21. 60 13 1. 28 2. 3. 45 4. Chapter 7 Test, Form B 1. 21 2. 3.5 3. 12.5 4. SLQ NTR by SSS Theorem 5. MBD FWY by AA Postulate 6. PRG KRN by SAS Theorem 7. not similar 8. 9.6 m 9. 15 10. B 11. 9 12. 6.4 13. 3 : 1 14. 2 : 3 15. x = 27 16. x = 10; y = 4.5 17. 4.8 Alternative Assessment, Form CTASK 1: Scoring Guide AA Similarity Postulate and SAS and SSS Similarity Theorems; check students’ work. 3 Student gives correct answers and accurate drawings and explanations. 2 Student gives mostly correct answers, drawings, and explanations. 1 Student gives answers, drawings, andexplanations that contain significant errors. 0 Student makes little or no effort. Geometry Chapter 7 All rights reserved. 1. ABCD is a trapezoid 2. AD 6 BC 3. AED CEB 4. EBC EDA 5. AED CEB 2. Statements 1. T, U, and V are mdpts 2. TU 6 RV, UV 6 TR, and TV 6 US 3. TUVR and TUSV areparallelograms. 4. TUV R Reasons 1. Given 2. Def. of a trapezoid 3. Vertical angles are . 4. Alt. int. angles are . 5. AA Similarity Post. Reasons 1. Given 2. Triangle Midsegment Theorem 3. Def. of parallelogram © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. Statements Activity 4:Thinking Chapter 7 Answers (continued) Claim 1 is true, and Claim 2 is false. Claim 3 is partly true; the triangles are similar, but the ratio of the areas is 1 : 4, not 1 : 2. Because the ratio of heights is not close to the ratio of head diameter, the baby and the adult are not similar. (The baby’s head will growmore slowly than the rest of the body.) 3 Student gives correct answers and provides logical reasons to support the answers. 2 Student gives mostly correct answers, but the work may contain minor errors or omissions. 1 Student gives answers that contain significant errors. 0 Student makes little or noeffort. 3 Student gives a mathematically sound analysis. 2 Student gives an explanation that contains minor errors or omissions. 1 Student gives an explanation containing significant errors or omissions. 0 Student makes little or no effort. TASK 3: Scoring Guide Cumulative Review All rights reserved.TASK 2: Scoring Guide One method is to pick any point A on land and measure AB and AC. Then select a fraction (say, 12), and mark X and Y halfway from A to B and C, respectively. Then measure XY. Then BC = 2 ? XY. ? C B X Y 3 Student gives a mathematically correct procedure and explanation.A 2 Student gives a procedure and explanation that may contain minor errors. 1 Student gives an invalid procedure or a valid procedure with little or no explanation. 0 Student makes little or no effort. 1. D 7. D 13. D 2. G 3. A 4. G 5. C 6. H 8. F 9. B 10. J 11. B 12. H 14. F 15. 10 16. Congruent triangleshave congruent corresponding angles and congruent corresponding sides, so they have a similarity ratio of 1 : 1. Similar triangles have congruent corresponding angles but do not necessarily have congruent corresponding sides. 17. Sample: 18. If cows do not have six legs, then pigs cannot fly. © PearsonEducation, Inc., publishing as Pearson Prentice Hall. TASK 4: Scoring Guide Comparing the adult to the baby, we have: • ratio of heights 3.5 : 1 • ratio of head diameter 1.7 : 1 Geometry Chapter 7 Answers 31 Chapter 8 Answers Practice 8-1 Practice 8-6 1. " 51 2. 4 " 7 3. 2 " 65 4. " 7 5. 2 " 21 6. 18 " 2 7.46 in. 8. 78 ft 9. 279 cm 10. 19 m 11. acute 12. right 13. obtuse 14. right 15. obtuse 16. acute Practice 8-2 1. x = 2; y = " 3 3. 6. All rights reserved. 14. w = b=3 6 4 2. a = 4.5; b = 4.5" 3 3 25"3 c = 50" 4. 8 " 2 5. 3 ;d = 3 28"3 7. 2 8. x = 15; y = 15 " 3 3 10. 42 cm 1. 46.0, 46.0 2. -37.1, -19.7 3. 89.2, -80.3 4.38.6 mi/h; 31.2º north of east 5. 134.5 m; 42.0º south of west 6. 1.3 m/sec; 38.7º south of east 7. 55º north of east 8. 20º west of south 9. 33º west of north y 10a. 1, 5 10b. 11. 5.9 in. 12. 10.4 ft, 12 ft 10"3 3 ;x = 14 " 2 9. 3 " 2 13. 170 in.2 3 5; y = 5 " 2 ; z = 5" 15. a = 4; 3 16. p = 4 " 3; q = 4 " 3; r = 8; s = 4 "6 2 4 6x 642 2 4 6 11a. 1, -1 y 11b. 6 4 Practice 8-3 1. tan E 3. tan E 7. 7.1 12. 71.6 17. 39 = 34 ; tan F = 34 = 52; tan F = 52 2. tan E = 1; tan F = 1 4. 12.4 5. 31.0 6. 14.1 8. 2.3 9. 6.4 10. 78.7 11. 26.6 13. 39 14. 72 15. 27 16. 68 18. 54 12a. -7, -1 y 12b. 6 4 2 Practice 8-4 1. sin P = 2"710 ; cos P = 5 12 13 ;

cos P = 13 "2 "2 = 2 ; cos P = 2 3. sin P = © Pearson Education, Inc., publishing as Pearson Prentice Hall. 5. sin P 7. 64 13. 6.6 3 7 8. 11.0 3 5 42 4. sin P = "611 ; cos P = 9. 7.0 14. 37 2. sin P = 54; cos P = 15. 11.0 6. sin P = 10. 42 16. 56 15 17 ; cos P 11. 7.8 4 6x 642 2 4 6 5 6 8 = 17 2 4 6x 4 6 12. 5317. 11.5 13. Sample: N 18. 9.8 Practice 8-5 1a. angle of depression from the birds to the ship 1b. angle of elevation from the ship to the birds 1c. angle of depression from the ship to the submarine 1d. angle of elevation from the submarine to the ship 2a. angle of depression from the plane to the person 2b.angle of elevation from the person to the plane 2c. angle of depression from the person to the sailboat 2d. angle of elevation from the sailboat to the person 3. 116.6 ft 4. 84.8 ft 5. 46.7 ft 6. 31.2 yd 7. 127.8 m 8. 323.6 m 9a. W 48 E S 14. Sample: N 30 W E S 35 30 ft 9b. 26 ft Geometry Chapter 8 Answers33 (continued) Reteaching 8-1 Enrichment 8-1 1.–2. Sample: Friar Tuck is the winner. By using the Pythagorean Theorem, Friar Tuck’s string length is 4 " 2 + 2 " 5 or 10.1 in. Little John’s string length is 4 " 5 + 2 " 2 or 11.8 in. Enrichment 8-2 1 Reteaching 8-2 1 1.–2. Sample: 1 1 7 1 8 1 9 or 3 10 11 1 1 13. Sample: 3.5 and 7 4. Sample: 3.5 and 7 5. Sample: They are approximately equal. 6. Check students’ work. 7. d = 2 " 3; e = 4 4"3 8"3 8. f = 3 ; g = 3 9. x = 3 " 3; y = 3 10. u = 6 " 3; v = 12 1. 3 2. –2 3. 45° 4. 135°; definition of supplementary angles 5. 45°; vertical angles are congruent. 6. 135°; verticalangles are congruent. 7. y x 1. DF = 2.7, EF = 7.5 2. XZ = 8.7, YZ = 10 3. RS = 8.4, ST = 13.1 4. 5.8 5. 30; 60 6. 16.6; 7. 23.6; 66.4 8. 2.1; 3.4 9a. 7.1 9b. "51 73.4 Reteaching 8-5 3. 2.2° 8. -2, 12 9. 90°, 90° 10. No; because the slopes of the lines are opposite reciprocals, the lines must be perpendicular.11. 70.6°, 109.4° 4. 42.0 ft Reteaching 8-6 1. 12.8 mi/h at 51.3° off directly across 2a. 10 mi 2b. 36.9° west of due north 3. The bird’s path shifts 31.0° east of due south. 4. 4 mi/h Answers 2y 4 Enrichment 8-4 1. 5.8 cm 2. 3.4 cm 3. 7.5 cm 4. 47 5. 25.5 6. 107.5 7. 0.731 8. 0.431 9. 0.951 10. 7.9 11. 7.9 12.7.9 13. They are equal. 14. The sides of a triangle are proportional to the sides of their opposite angles. 15. 70 16. 15.55 km 17. 11.83 km 18. Lighthouse Q; it is closer. Geometry Chapter 8 © Pearson Education, Inc., publishing as Pearson Prentice Hall. x Reteaching 8-4 34 1 Enrichment 8-3 1. 26.6 2.76.0 3. 80.5 4. 26.6 5. 8.1 6. 0.8 7. mA = 53.1; mB = 36.9 8. mA = 67.4; mB = 22.6 9. mA = 70.5; mB = 19.5 10. mA = 63.4; mB = 26.6 11. 26.6 12. 63.4 13. 49.4 2. 1113.8 ft 6. 19.1 ft 1 1 y 2x 3 Reteaching 8-3 1. 42.9 ft 5. 76.1 ft 4. "5 14 15 sum of the squares of the two legs equals the square of the 6. 4 "5 7. 7 " 2 8. 8 9. 6 hypotenuse. 1 4 or 5 2 6 13 5. The 1 1 2 4. They are equal. 3. " 4 or 2 1 3. Sample: 9, 16, and 25 2. " 3 3 2 1. " 2 5. All rights reserved. Chapter 8 Answers Chapter 8 Answers (continued) Chapter Test, Form A Enrichment 8-5 1. the top of the tower 3. P 1. right 2. PM 2. acute 6. x 5 7Í 2;y 5 7Í 2 P 8. sin A All rights reserved. B x x 9. sin A 60 y M 45 A y 100 M 11. 20.9 13. 26.6 4. the angle of elevation from B to the top of the tower; 5. the angle of elevation 30°-60°-90°; scalene right triangle from A to the top of the tower; 45°-45°-90°; isosceles right x x 6. tan 60 = y 7. tan 45 = y 1 100triangle 8. x = tan 60(y); x = tan 45(y + 100) 9. 136.6 10. 236.6; the height of the tower 11. yes 1. (0.5, 0.5) 2. (-0.25, 1.80) 4. 2.9 km; 38.7° south of east 3. (-2.25, 1.80) Chapter Project 67 elevation from the top of the tree to the birds y as the side adjacent to Y. The side adjacent Z to X is the same as theside opposite Y. Y adjacent /Y So, tan x° = adjacent /X = opposite /Y = tan1 y8 . 18. 92.9, 83.6 19. 120.4, -258.3 20. 211.9 mi; 21. 11, 0 22. 4, 3 19° south of west 23. Sample: y 4 2 4 2 2 line of sight 2 4 x 4 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 7. x 5 4; y 5 4Í 3 14d. angle ofdepression from the birds to the x top of the tree 15. 102 ft 16. 143 ft 17. Sample: The side opposite X is the same 1 and 2 are complementary, and 2 and 3 are complementary, so m1 = m3. horizon line 2 3 Chapter Test, Form B edge of protractor 1. right 5. sin T Activity 2: Measuring 6. sin T 8. 15.3Check students’ work. Check students’ work. Activity 4: Researching Check students’ work. ✔ Checkpoint Quiz 1 2. right = 54, cos A 3. acute = 53; tan B = 34, sin 4. tan A = 34, B = 35, cos B = 45 5. tan A = 2, sin A = 0.8955, cos A = 0.4478; tan B = 0.5, sin B = 0.4478, cos B = 0.8955 6. tan A = 0.8833,sin A = 0.6625, cos A = 0.75; tan B = 1.1321, sin B = 0.75, cos B = 0.6625 7. 11.6 8. 41° 9. 16.5 ✔ Checkpoint Quiz 2 1. (71, 79) 2. (16, -9) 3. (-60, 148) north of east 5. 16.1, 29.7° south of west 48.4° north of west 7. 22.5 ft Geometry Chapter 8 2. obtuse 3. 2Í 85 4. x 5 4Í 3; y 5 8Í 3 6 "2 6 "2 = 19 ; cos T =17 19 ; tan T = 17 93 14 = 17 ; cos T = "1793; tan T = 14" 93 7. 60 12a. angle of 9. 63 10. 36.9 11. 78.7 depression from the cloud to the person in the castle 12b. angle of elevation from the person in the castle to the 12c. angle of depression from the person in the cloud 12d. angle of elevation castle to theperson on the ground 13. 1710 ft from the person on the ground to the castle Activity 3: Comparing sin A 5. 3 12. 76.0 14a. angle of elevation from the person to the top of the tree 14b. angle of depression from the top of the tree to the person 14c. angle of X Activity 1: Building 1. obtuse 4. 2Í 41 2"42 42 19= 23 ; cos A = 23 ; tan A = 2" 19 3 "7 "7 = 43 ; cos A = 4 ; tan A = 7 10. opposite /X Enrichment 8-6 string 3. obtuse 4. 31.6, 71.6° 6. 120.4, 14. sin x° = opposite hypotenuse adjacent and cos x° = hypotenuse opposite adjacent sin x8 cos x8 = hypotenuse hypotenuse = X x opposite hypotenuse hypotenuse? adjacent = opposite adjacent = 15. -92.3, 59.9 17. 1, 10 tan x° y Z Y 16. 18.0 mi; 33.7° north of east Answers 35 Chapter 8 Answers (continued) 18. Sample: TASK 4 Scoring Guide: a. a 12, -3; b 8, 8 b. a 12, 4; b 11, 3 c. a 14° south of east; b 45° north of east W W y 4 4 2 2 4 x 4 3 2 1 0 W W W WStudent gives correct answers. Student gives answers that are mostly correct. Student gives mostly incorrect answers. Student makes little or no effort. x = 57.3 1. A 2. H 3. D 4. G 5. D 6. G 7. C 8. J 9. D 10. F 11. B 12. J 13. D 14. 10 15. Sample: You may have used the branches to climb into the tree. 16.Sample: Suppose that an 3 Student gives correct answer and shows a valid method. 2 Student gives generally correct work that contains minor errors. 1 Student gives an incorrect answer, and the method is not correct. 0 Student makes little or no effort. equilateral triangle has an obtuse angle. Then oneangle in the triangle has measure greater than 90. This contradicts the fact that, in an equilateral triangle, the measure of each angle is 60. 17. Sample: The side opposite the 90° angle is also the hypotenuse, and there are two adjacent sides. 18. Sample: The sum of the three angles of a triangle is 180°.On a sphere, the sum of the three angles will be greater than 180°. Alternative Assessment, Form C TASK 1 Scoring Guide: All rights reserved. Cumulative Review TASK 2 Scoring Guide: 111 ft © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3 Student gives a correct answer and shows avalid method. 2 Student gives generally correct work that contains minor errors. 1 Student gives an incorrect answer, and the method is not correct. 0 Student makes little or no effort. TASK 3 Scoring Guide: x = 1.53 mi; y = 0.52 mi 3 Student gives correct answer and shows a valid method. 2 Student givesgenerally correct work that contains minor errors. 1 Student gives an incorrect answer, and the method is not correct. 0 Student makes little or no effort. 36 Answers Geometry Chapter 8 Chapter 9 Answers 9. Practice 9-1 1. No; the triangles are not the same size. All rights reserved. 2. Yes; the hexagonsare the same shape and size. 3. Yes; the ovals are the same shape and size. 4a. C and F 4b. CD and CD, DE and DE, EF and EF, CF and CF 5a. M and N 5b. MN and MN, NO and NO, MO and MO 6. (x, y) S (x - 2, y - 4) 7. (x, y) S (x - 2, y - 2) 8. (x, y) S (x - 3, y - 1) 9. (x, y) S (x + 4, y - 2) 10. (x, y) S (x -5, y + 1) 11. (x, y) S (x + 2, y + 2) 12. W(-2, 2), X(-1, 4), Y(3, 3), Z(2, 1) 13. J(-5, 0), K(-3, 4), L(-3, -2) 14. M(3, -2), N(6, -2), P(7, -7), Q(4, -6) 15. (x, y) S (x + 4.2, y + 11.2) 16. (x, y) S (x + 13, y - 13) 17. (x, y) S (x,y) 18. (x, y) S (x + 3, y + 3) 19a. P(-3, -1) 19b. P(0, 8), N(-5, 2), Q(2, 3) A Practice 9-2 1. (-3, -2)2. (-2, -3) 3. (-1, -4) 4. (4, -2) 5. (4, -1) 6. (3, -4) y 7a. y T B 4 2 6 4 2 2 2 4 B T A 10. T y T 4 6 4 2 A 11. 6 x 4 2 4 4 B B A y T 6 x T 4 4 L 2 6 4 2 O 2 I 4 6 x K J B A 4 A y B 12. y 7b. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 8 6 4 2 A 2 6 8 x B I J O 2 6 L 2 4 6 6 x 2 6 4 2 2 y 4 Z A2 Y 6 4 2 2 4 W 4 8b. 6x X 13. (-6, 4) 1. I 7. 2. I 4 6x 14. (-8, 0) 3. I 15. (0, -12) 4. GH 5. G 6. ST U T P 4 Y 2 2 4 B Practice 9-3 y X T, T 4 4 8a. 6 4 4 10 4 K 2 W Z 4 Geometry Chapter 9 4 6x S T U S Answers 35 Chapter 9 Answers (continued) 8. Practice 9-4 1. The helmet has reflectional symmetry. 2.The teapot has reflectional symmetry. 3. The hat has both rotational and reflectional symmetry. 4. The hairbrush has P reflectional symmetry. 5. O 9. P P N Q O 6. Q O P N Q M N L L Q O L M N L M N M L O Q Q P line symmetry and 72° rotational symmetry P 9. N O 11. L 12. E 7. This figure has no linesof symmetry. 8. P Q M P O line symmetry Q' 10. P' S R' P line symmetry and 90° rotational symmetry Q 11. R 13. P' S Q' line symmetry and 45° rotational symmetry P R' 12. Q X O X R 180° rotational symmetry 13. line symmetry 36 Answers Geometry Chapter 9 © Pearson Education, Inc., publishing asPearson Prentice Hall. 10. All rights reserved. N Chapter 9 Answers (continued) 14. 12. R R line symmetry and 45° rotational symmetry O 15. A 180° rotational symmetry A All rights reserved. 16. T 13. P(-12, -12), Q(-6, 0), R(0, -6) 14. P(-12, 41), 3 1 1 Q(1 4, -4), R(1 4 , 2) 15. P(-21, 6), Q(3, 24), R(-6, 6)16. P(-2, 1), Q(-1, 0), R(0, 1) Practice 9-6 line symmetry 17. 18. 1. I. D II. C III. B IV. A 3. COOK 2. I. B II. A III. C IV. D B B B 20. H O A X m m 4. C C C 19. 21. m 5. 1. L(-2, -2), M(-1, 0), N(2, -1), O(0, -1) 2. L(-30, -30), M(-15, 0), N(30, -15), O(0, -15) 3. L(-12, -12), M(-6, 0), N(12, -6), O(0, -6) 4. 35 5. 12 6. 27. yes 8. no 9. no R A R 11. R O T A m J Practice 9-5 10. J R Geometry Chapter 9 A m y 6. 4 E S 2 T A J © Pearson Education, Inc., publishing as Pearson Prentice Hall. T T T O 2 2 4 6 x B Answers 37 Chapter 9 Answers y y 13. 10 B T 4 B 8 6 4 2 O 2 4 E S 2 4 2 O 4 x 2 4 2 4 8 x 6 B T translationalsymmetry y T 2. S 4 2 2 O B 2 16. glide reflection 1. 4 line symmetry, rotational symmetry, translational symmetry, glide reflectional symmetry E 2 6 x 4 3. y 10. 4 B T 2 translational symmetry 6 4 2 O 2 E 4 4 x 2 4. line symmetry, rotational symmetry, translational symmetry, glide reflectional symmetry S y11. B 2 5. 4 x 6 E 2 O 2 line symmetry, rotational symmetry, translational symmetry, glide reflectional symmetry 4 S T 6. y 12. S 4 E 2 2 O 2 T 38 2 4 4 rotational symmetry, translational symmetry 6 x 7. 8. B Answers Geometry Chapter 9 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 9.S Practice 9-7 2 2 O 2 6 x 4 15. rotation 14. reflection 17. translation E S 2 4 E y 8. T 2 All rights reserved. 7. (continued) Chapter 9 Answers (continued) 9.–11. Samples: 9. 10. 8. Z Y X 11. Z X All rights reserved. Reteaching 9-4 12. yes 17. no 13. yes 14. no 15. yes 16. no Reteaching 9-1 1.–5. Checkstudents’ work. 6. A(0, -3), B(1, 1), C(4, -1), D(5, -4) 7. A(4, -2), B(5, 2), C(8, 0), D(9, -3) 8. A(3, 5), B(4, 9), C(7, 7), D(8, 4) 1. two lines of symmetry (vertical and horizontal), 180° rotational symmetry (point symmetry) 2. one line of symmetry (horizontal) 3. one line of symmetry (vertical) 4. two lines ofsymmetry (vertical and horizontal), 180° rotational symmetry (point symmetry) 5. one line of symmetry (vertical) 6. one line of symmetry (vertical) Reteaching 9-5 1. Check students’ work. y 2. 4 Reteaching 9-2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1.–5. Check students’ work. 6.reflection across x-axis: F(-1, -3), G(-5, -1), H(-3, -5); reflection across y-axis: F(1, 3), G(5, 1), H(3, 5) 7. reflection across x-axis: C(2, -4), D(5, -2), E(6, -3); reflection across y-axis: C(-2, 4), D(-5, 2), E(-6, 3) 8. reflection across x-axis: J(-1, 5), K(-2, 3), L(-4, 6); reflection across y-axis: J(1, -5), K(2, -3), L(4, -6) 2 4 2 O 2 2 4 x 4 y 3. 4 Reteaching 9-3 2 Y 1.–5. 4 Z Y X X 2 4 x 4 Z y 4. T Y 6. O 2 4 2 Z X X Y T Z 12 8 6 4 2 O 2 2 4 6 8 12 x 4 7. Sample: Y Z S X X Y Z Geometry Chapter 9 Answers 39 Chapter 9 Answers 12 10 y 5. line symmetry across the dashed lines, rotational symmetry around points,translational symmetry, glide reflectional symmetry 6 4 2 8 6 2 2 4 6 8 x Enrichment 9-1 4 6 8 10 1. 3. 5. 8. (x, y) S (x - 7, y - 2) 2. (x, y) S (x + 4, y - 7) (x, y) S (x - 10, y + 5) 4. (x, y) S (x - 3, y - 4) Foster 6. yes 7. Wilson (x, y) S (x - 7, - 3) 9. C 12 Enrichment 9-2 1. (-3, -1) Reteaching 9-6 1. translation 2.reflection 3. rotation 4. glide reflection 5. rotation 6. glide reflection 7. reflection 8. translation 2. (-1, 0) 3. (5, 3) 4. the midpoint formula: M = y 5. x1 1 x2 y1 1 y2 ( , 2 2 ) 6 L K 4 Reteaching 9-7 2 1. L' J 6 4 2 J' All rights reserved. 5. (continued) 0 2 4 6 x 2 4 K' 2. Sample: 6. y = 21 x + 7. 1 2 T' line symmetryacross the dashed lines, rotational symmetry around points, translational symmetry, glide reflectional symmetry 6 4 2 Q' 4 2 0 6x R' R 6 Q 8 line symmetry across the dashed lines, rotational symmetry around points, translational symmetry, glide reflectional symmetry 2 S'4 2 4 3. y S T y = -4 4. rotationalsymmetry around points, translational symmetry 40 Answers Geometry Chapter 9 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 6 Chapter 9 Answers 8. 6 y (continued) 12. E' 6 D' 4 4 D 2 H' 2 F' 0 6 4 2 2 4 8 6 4 2 0 H 2 6 I G 4 2 J' 4 6x J 6 y = -x K 9. 6 All rights reserved. I' K' 6x 2 F 4E y y 4 A D A' 6 4 0 B 2 C 4 2 C' Enrichment 9-4 x=1 10. M 6 y N M' 2 P 0 8 6 4 2 N' Q 2 Q' 4 P' 6x 4 6 y = -2x - 2 11. 1 1. (0, -2, 3) 2. (-2, -2, 3) 3. (-2, -2, 0) 4. (0, -2, 0) 5. (0, -6, 0) 6. (0, -6, 3) 7. (-2, -6, 3) 8. (-2, -6, 0) 9. (2, 0, 3) 10. (2, -2, 3) 11. (2, -2, 0) 12. (2, 0, 0) 13. (6, 0, 0) 14. (6, 0, 3) 15. (6, -2, 3)16. (6, -2, 0) x 6 B' = -12 x - Enrichment 9-3 D' 2 6 © Pearson Education, Inc., publishing as Pearson Prentice Hall. y G' U' V' 6 y U 4 Enrichment 9-5 2 W' T' 8 6 4 2 0 2 4 6 y = 3x + 2 Geometry Chapter 9 1. yes; rotational and point symmetry 2. no 3. yes; rotational and point symmetry 4. yes; rotationaland point symmetry 5. no 6. no 7. diamonds 8. 12 9. Seven of diamonds; this does not have symmetry because the diamond in the middle is toward either the bottom or top of the card, and when you rotate the card 180°, the position will be reversed. 10. All the face cards have symmetry. 11. yes; 2, 4, 1012. No; Sample: When you look at the card one way, three of the points of the hearts are pointing down, and five are pointing up. When you rotate the card 180º, five of the points of the hearts are pointing down, and three are pointing up. 13. No; because the number and suit of each card are placed inopposite corners, none of the cards have line symmetry. 14. You can add a backward 3 with the small club below it to the two empty corners to create line symmetry, or you can remove the 3 with the small club below it from each of the two corners. V 6x 4 T W 1a. (2, 0, 2) 1e. (2, 0, 0) 2a. (4, 0, 4) 2e. (4, 0,0) 3. (0, 0, 0) 1b. (0, 0, 2) 1c. (0, 2, 2) 1d. (2, 2, 2) 1f. (0, 0, 0) 1g. (0, 2, 0) 1h. (2, 2, 0) 2b. (0, 0, 4) 2c. (0, 4, 4) 2d. (4, 4, 4) 2f. (0, 0, 0) 2g. (0, 4, 0) 2h. (4, 4, 0) 4. 2 5. Each edge in the image is double that in the preimage. 6. Samples: Three faces are coplanar for image and preimage; three faces areparallel; each face in the image has two times the perimeter and four times the area of the corresponding face in the preimage. 7. Surface area of image = 96 sq. units; surface area of preimage = 24 sq. units; the ratio of the surface areas is 4 : 1. Answers 41 Chapter 9 Answers (continued) 8. Volume ofimage = 64 cubic units; volume of preimage = 8 cubic units; the ratio of the volumes is 8 : 1. 9. Dilations in three dimensions are proportional to dilations in two dimensions. 1.–9. Check students’ work. 10. 12-3; rotation; twice 11. 130 12. 100 13. 160 7. 8. C C' L' B B' B D' A B' D C C' A, A' 9. L C' C A D' D12 R Y O T A T I O N 10. T( 8, -5), Q(2, -3) and R(4, 0) ✔ Checkpoint Quiz 2 1. line symmetry; rotational; 180° H 2. line symmetry; rotational; 180° 3. line symmetry Activity 2: Modeling a. Ancient Egyptian ornament b. Oriental design 4. rotational; 180° 5. translational; no turns 6. rotational; preimage turns7. translational 8. rotational, reflectional, translational 9. translational Chapter Test, Form A c. Greek vase design 42 Answers All rights reserved. 11 units up 1. 2. 3. 4. 5. P(13, 0), Q(10, -2), R(11, -4), S(13, -6) P(-1, -2), Q(2, -4), R(1, -6), S(-1, -8) P(0, -5), Q(2, -2), R(4, -3), S(6, -5) P(-2.5, 0), Q(-1, -1), R(-1.5, -2), S(-2.5, -3) P(5, 2), Q(2, 0), R(3, -2), S(5, -4) 6. P(-11, 3), Q(-8, 1), R(-9, -1), S(-11, -3) 7. glide reflection 8. translation 9. translation 10. rotation Geometry Chapter 9 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Activity 1: Investigating a. vertical b. vertical c. both Check students’work. 1. No; the figures are not the same size. 2. yes 3. yes 4. (x, y) S (x - 3, y - 2) 5. 6 units right and 4 units down 6. a resultant translation of 4 units right and 2 C 3 O R 5 4 M P E G P R F L O E L I 6 I S E D I M I E C S 7 R O T A T I O N A L G I I E M E O O F E N N L T 10 A E R R 11 C L S Y M M E T13 P O I N T F I L 14 15 E T R A N S F O R M A T I O N N C I A T L 16 I T E S S E L L A T I O N N O G N Chapter Project Activity 5: Creating ✔ Checkpoint Quiz 1 Enrichment 9-7 T R A N S L A T 8 I M A G O N 9 L I N a, b Activity 4: Classifying a. mg b. 12 Enrichment 9-6 1 Activity 3: InvestigatingChapter 9 Answers (continued) 11. 16. B' C' A' P A C All rights reserved. 12. line symmetry 13. rotational symmetry 14a. Any two of the following: ABCDEHIKMOTUVWXY 14b. Any two of the following: HINOSXZ 15. Translations can be performed on the pieces because they can slide. Rotations can beperformed on the pieces because each piece can be turned. Reflections and glide reflections cannot be performed on the pieces because the back of a puzzle piece is useless in solving the puzzle. Dilations are impossible because the pieces cannot change size. 16. A 17. Sample: glide reflectional,rotational, line, and translational © Pearson Education, Inc., publishing as Pearson Prentice Hall. 18. L(-3, 6), M(6, 6), N(3, -9) 19. L(12, -34), 3 5 1 M(4, 1), N(4, -2) 20. L(-18, -18), M(-18, 18), N(0, 0) 21. L(3, -2), M(0, -53 ), N(73 , -2) 22. YZ = 15 cm; XZ = 24 cm; scale factor = 73 23. XY = 25 in.; YZ = 40in.; scale factor = 53 24. (-1, 3) 25. (5, 4) 26. (-2, -7) 27. (2, -7) 28. (1, -1) 29. (-3, 1) B 17. 180° rotational symmetry 18. line symmetry 19. line symmetry; 180° rotational symmetry 20. (-7, 0) 21. (0, -3) 22. (-6, 1) 23. (-9, -4) 24. (2, -2) 25. (-5, 11) Alternative Assessment, Form C TASK 1: Scoring Guide 20;200 cm; all distances are magnified 20 times. Chapter Test, Form B 1. Yes; the figures are the same shape and size. 2. No; the figures are not the same size. 3. Yes; the figures are the same shape and size. 4. (x, y) S (x - 6, y + 1) 5. (x, y) S (x + 3, y - 7) 6. A9(0, -3), B9(-4, 0), C9(-6, -4), D9(-3, -5) 7. A9(6,3), B9(10, 0), C9(12, 4), D9(9, 5) 8. A9(3, 8), B9(-1, 5), C9(-3, 9), D9(0, 10) 9. A9(5, 2), B9(1, -1), C9(-1, 3), D9(2, 4) 10. A9(-3, 0), B9(0, -4), C9(-4, -6), D9(-5, -3) 11. A9(0, 3), B9(4, 6), C9(6, 2), D9(3, 1) 12. (x, y) S (x + 4, y - 3) 13. (x, y) S (x - 3, y + 6) 14. rhombus and two triangles; 3 Student gives correctanswers and a clear and accurate explanation. 2 Student gives answers and an explanation that are largely correct. 1 Student gives answers or an explanation that may contain serious errors. 0 Student makes little or no effort. TASK 2: Scoring Guide Sample: translation 15. A' B' P C' A C B GeometryChapter 9 3 Student draws a figure that accurately reflects the stated conditions. 2 Student draws a figure that is mostly correct but falls just short of satisfying all the stated conditions. 1 Student draws a figure that has significant flaws relative to the stated conditions. 0 Student makes little or no effort.Answers 43 Chapter 9 Answers (continued) d. glide reflection: TASK 3: Scoring Guide Sample: The car can undergo translations and rotations as it moves and turns. But it can undergo neither a dilation, which would change its size, nor a reflection, which would, for example, put the steering wheel on thepassenger side. 3 Student gives a correct and thorough explanation. 2 Student gives an explanation that demonstrates understanding of transformations but may contain minor errors or omissions. 1 Student gives an explanation containing major errors or omissions. 0 Student makes little or no effort. 4 yThe glide reflection is equivalent to a translation 2, 0 followed by a reflection across the line y = -2. 2 4 2 2 2 4x 4 e. rotation: 4 y 4x 4 2 4 y 4 2 3 Student gives accurate graphs and correctly identifies the details the problem calls for. 2 Student gives generally correct graphs and details, although the workmay contain some errors. 1 Student gives graphs and accompanying details that contain significant errors or omissions. 0 Student makes little or no effort. 4x 4 4 a. translation: 4 y 2 translation vector: 1, 0 4 2 2 Cumulative Review 1. D 8. H 14. J 2. G 3. B 4. H 5. A 6. J 7. B 9. C 10. J 11. B 12. H 13. A 15.B 16. 49 cm3 17. A, H, I, M, O, T, U, V, W, X, Y 18. Rotational; it is turning around a fixed point, 4x 4 the corner. b. dilation: 19. 4 4 y 2 T scale factor: 2 y U G U G 2 T 2 2 x 2 4 2 4 x 20. Equilateral triangle, square, and regular hexagon; the measure of each interior angle is a factor of 360. 21. If a triangleis not a right triangle, then the Pythagorean Theorem cannot be applied. c. reflection: y 2 4 2 2 2 line of reflection: y = 1 2 y 22. 4 4x 2 U 4 V V U 4 2 2 4 W 6 x W 4 23. 21.21 44 Answers Geometry Chapter 9 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Sample: All rights reserved.degrees of rotation: 90 TASK 4: Scoring Guide Chapter 10 Answers Practice 10-1 1. 8 2. 8 3. 60 4. 240 5. 2.635 6. 1.92 7. 0.8125 8. 9 9. 1500 10. 8.75 11. 3.44 12. 7 13. 110 14. 30 15. 16 16. 119 17. 12 18. 48 19. 40 20. 56 Practice 10-2 All rights reserved. 1. 48 cm2 2. 784 in.2 3. 11.4 ft2 4. 90 m2 2 2 25. 2400 in. 6. 374 ft 7. 160 cm 8. 176.25 in.2 2 9. 54 ft 10. 42.5 square units 11. 96 " 3 square units 12. 36 " 3 square units 13. 45 cm2 14. 226.2 in.2 2 15. 49,500 m Practice 10-3 8"3 8"3 = 3 ;d = 3 1. x = 7; y = 3.5 " 3 2. a = 60; c 3. p = 4 " 3; q = 8; x = 30 4. m1 = 45; m2 = 45; m3 = 90; m4 = 90 5. m5 = 60;m6 = 30; m7 = 120 6. m8 = 60; m9 = 30; m10 = 60 7. 12 " 3 8. 36.75 " 3 9. 75 " 3 10. 120 in.2 2 2 11. 137 in. 12. 97 in. 1. 4 : 5; 16 : 25 2. 5 : 3; 25 : 9 3. 3 : 4; 9 : 16 2 5. 9 : 5 6. 1 : 6 7. 8 : 3 8. 576 9. 5 in. 1152 2 10. 25 in. 11. 2 : 3 12. 1 : 2 13. 2 : 5 14. 108 ft2 4. 1 : 2 2700 7 cm2 17. p 2 18. 3 Practice 10-87 3 1. 8.7% 2. 10.9% 3. 15.3% 4. 10 5. 10 1 1 2 6. 5 7. 10 8. 1 9. 5 10. 22.2% 11. 4.0% 12. 37.5% 13. 20% 14. 50% 15. 40% 16. 33 13 % Reteaching 10-1 1. Samples: 4 4 6 4 6 6 2. Samples: 4 3 6 cm2 3. 20 4. 55 8. Samples: Practice 10-4 © Pearson Education, Inc., publishing as Pearson Prentice Hall.15. 63 16. 25 8p 3p 19. 2 20. 10 3 8 in.2 2 8 5. 6 m 6. 8 ft 7. 4 in. 4 12 3 6 8 9. 5 in. Practice 10-5 10. 23 cm 11. 9 ft 1. 174.8 cm2 2. 578 ft2 3. 1250.5 mm2 2 2 4. 192.6 m 5. 1131.4 in. 6. 419.2 cm2 7. 324.9 in.2 8. 162 cm2 9. 37.6 m2 10. 30.1 mi2 11. 55.4 km2 12. 357.6 in.2 13. 9.7 mm2 14. 384.0 in.2 15.54.5 cm2 16. 9.1 ft2 2 2 17. 80.9 m 18. 83.1 m 19. 65.0 ft2 20. 119.3 ft2 2 2 21. 8 m 22. 55.4 ft Reteaching 10-2 Practice 10-6 2. Sample: bases = 6 cm and 10 cm, height = 6 cm 3. Sample: 48 cm2 4. Check students’ work. 5. Sample: 1 1 1. 32p 2. 16p 3. 7.8p 4. Samples: DFA , ABF 00 1 1 5. Samples: FE, FD 6. Samples: FEB , EDA 0 0 7. Samples: FE , ED 8. ACB 9. FCD 10. 32 11. 43 12. 54 13. 39 14. 71 15. 90 16. 50 17. 220 18. 140 19. 220 20. 130 21. 6p in. 22. 16p cm 23. 49 p m Practice 10-7 1. 49p 2. 21.2 6. 81 7. or 9 18 10. 5 p 11. 8 p p 4 3. 49 6p p 1 16 p or 16 12. 23 p Geometry Chapter 10 4.49 6p p 8. 16 - 81 13. 15 2p 21.2 5. 9. 4p 14. 6p 1 4p 1. Sample: 6. Sample: diagonals = 4 cm and 9 cm 8. Check students’ work. 9. 144 ft2 2 2 11. 75 m 12. 98.4 cm 7. Sample: 18 cm2 10. 112 in.2 Answers 39 (continued) Reteaching 10-3 Enrichment 10-1 1.–2. Sample: 1. c; each segment is thehypotenuse of a right triangle with 2. No; the angles must be shown to be legs a and b. 3. They are congruent; SSS 4. 180 right angles. 5. 90; acute angles of a right triangle are complementary. 6. CPCTC 7. mSPQ = 180 - (mAPS + mPSA) = 180 - 90 = 90 8. Repeat steps similar to those in Exercises 4–7to show that PQR, QRS, and RSP are 9. a + b; (a + b)2 = a2 + 2ab + b2 right angles. 2 10. c; c 11. a, b; 21 ab 12. The areas of congruent 13. a2 + 2ab + b2 = 4(12 ab) + c2 triangles are equal. or a2 + 2ab + b2 = 2ab + c2 14. By subtracting 2ab from both sides, you obtain a2 + b2 = c2 (The PythagoreanTheorem). radius apothem 3. Sample: apothem = 2.5, radius = 3.5, side = 5 4. yes 5. They are approximately equal. 6. 100 in.2 7. 200 ft2 8. 72 cm2 9. 8 m2 Reteaching 10-4 1. 4 : 3; 16 : 9 2. 8 : 5; 64 : 25 4. 4 : 3 5. 4 : 5 6. 3 : 5 9. 9 : 25 3. 9 : 5; 81 : 25 7. 16 : 9 8. 16 : 25 Enrichment 10-2 Reteaching 10-51. 19.3 in.2 5. 363.3 in.2 2. 123.1 cm2 6. 126.3 in.2 3. 86.6 in.2 7a. 81 in.2 4. 172.0 in.2 7b. area of square = s2 Reteaching 10-6 1. Sample: 1. Choose an interior point X; construct PX 6 BC, XQ 6 AC , and XR 6 AB. 2. A triangle is a trapezoid with one side where b1 = 0. Hence A = 21 h(b1 + b2) = 12 hb2. 3. Area of BDE = Area ABEC - Area ABDC = [ 21 h(y + x + b)] - [ 21 h(y + x )] = 12 hy + 12 hx + 21 hb - (12 hy + 21 hx) = 12 hb All rights reserved. Chapter 10 Answers Enrichment 10-3 2. Sample: 135° 3. Sample: 2 cm 4. Sample: 3p 2 cm 4.7 cm 5. Sample: 4.5 cm 6. They are approximately 10p equal.7. 5p 8. 9. 5p 10. 20p 11. 5p 3 3 2 6 15p 25p 12. 6 13. 2 2. 47 square units 3. 73 square Enrichment 10-4 1. 14 2. 21 3. 43 4. 19 Enrichment 10-5 1. 36 cm2 2. 4 "3 cm or about 6.93 cm 4. A = 6 sin u, 0° u 180° Reteaching 10-7 1.–2. Sample: 3. 6 cm Enrichment 10-6 4 cm 1.–5. Object 3. Sample: 50 cm2cm2 4. Sample: 50.3 5. They are approximately equal. 6. Check students’ work. 7. 25p ft2 8. 4p in.2 9. 64p m2 10. 20.25p ft2 2 2 11. 324p cm 12. 64p in. 13. 14 cm 14. 615 cm2 1. Quarter 2. CD 3. Circle with a radius of 4 in. 4. 5. Radius (r) 1.15 cm 238 in. 4 in. Diameter Circumference (d) (C) 2.3 cm 4 34in. 8 in. 7.222 cm 14.92 in. 8p Check student’s work. Check student’s work. 6. They are approximately equal. 8. C d 3.14 3.14 p 3.14 3.14 7. p 3.14 Reteaching 10-8 1. Check students’ work. 2. 21% 3. Check students’ work. 4. Sample: They are approximately equal. 5. Sample: The ratio would tend to beclose to 21%. 6. 43% 40 Answers Geometry Chapter 10 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. 19.5 square units units Chapter 10 Answers (continued) Enrichment 10-7 3 3 1. p 2. " 3 3. 3" 4. p - 3" 4 4 6. p4 7. 4 8. 1 9. p 10. " 2 12. p - 2 Enrichment 10-8 1. 0.06 or 1 18 2. 0.413. Alternative Assessment, Form C 5. 21 11. 2 0.63 or "3 1 1 "3 b h 36 in.2; 30.25 in.2; the seams in the finished quilt block account for the difference. b A parallelogram with base b and height h has the same area as two triangles with base b and height h. Thus, the area of the parallelogram is: A = 2 (12bh) = bh. b. b1 Activity 2: Creating h Check students’ work. b2 Activity 3: Researching Check students’ work. ✔ Checkpoint Quiz 1 1. 37.2 m2 5. 200 ft2 2. 20 ft2 6. 648 m2 h 4. 0.21 Activity 1: Modeling All rights reserved. Sample: a. Chapter Project 3. 63 m2 4. 96 in.2 7. 37; 949 ✔ Checkpoint Quiz 2 ©Pearson Education, Inc., publishing as Pearson Prentice Hall. TASK 1: Scoring Guide 1. 119.3 ft2 2. 293.9 m2 3. 523.1 in.2 4. 6.9 in. 2 5. 20.9 ft 6. 10p mi; 25p mi 7. 18p cm; 81p cm2 p or about 21% 8. 4 2 9. 12 or 50% 4 A trapezoid with bases b1 and b2 and height h has the same area as two triangles.The bases of the two triangles are b1 and b2, and the heights of both are h. Thus, the area of the trapezoid is: A = 21 b1h + 12 b2h = 12 h(b1 + b2). 3 Student gives accurate explanations accompanied by clear and correct diagrams. 2 Student gives generally correct explanations, although there may besome errors in either the explanations or the diagrams. 1 Student gives inaccurate or incomplete explanations and diagrams. 0 Student makes little or no attempt. Chapter 10 Test, Form A TASK 2: Scoring Guide 1. 15.6 ft2 2. 41.6 cm2 3. 192 ft2 4. 80 square units 5. 120 square units 6. 50 square units 7.64 8. 91 9. D 10. 18.8 cm2 11. 19.1 in.2 9 12. 268.4 ft2 13. 166.3 in.2 14. 172.0 cm2 15. Sample: Draw an altitude to form a 30°-60°-90° triangle. 16. 13.5 cm The altitude is 4" 3, and the area is 32 " 3. 17. 20 square units 18. 12p 19. 36p 20. 45 p 21. 54 p 22. p5 23. 4.44 square units 24. 21.5% 25. 9.1%Sample: C(2, 8), D(-2, 5). The slopes of AB and CD are equal, and the slopes of BC and AD are equal, so ABCD is a parallelogram. The slopes of AB and BC are opposite reciprocals, so AB # BC, and ABCD is a rectangle. AB = 5 and BC = 5, so the area of ABCD is 25 square units. Chapter 10 Test,Form B TASK 3: Scoring Guide 1. 110.9 ft2 2. 649.5 cm2 ; 649.6 cm2 ; or 649.7 cm2 3. 40.2 ft2 4. 60 ft2 5. 92 cm2 6. 72 in.2 16 1 2 7. 81 8. 4 9. A 10. 72.7 in. 11. 42.2 cm2 12. 21.6 cm 13. 385 cm2 ; 386 cm2 ; or 387 cm2 14. 36p 15. 324p 16. 4p 17. 18p 18. 75.4 cm2 19. 144p cm2 20. 4p in.2 GeometryChapter 10 3 Student gives correct answers and provides complete and accurate reasoning. 2 Student gives generally correct answers and reasoning. 1 Student gives generally incorrect answers and reasoning. 0 Student makes little or no attempt. Q Sample: 1 a. QXY is a 45°-45°-90° triangle, so theapothem QY = XY = 1 , or "22. "2 X Y The perimeter is 4(2XY) = 4(2"22) = 4 " 2 . The area is (2XY)2 = 4 ? ("22)2 = 4 (24) = 2. Answers 41 Chapter 10 Answers (continued) b. PAB forms a 30°-60°-90° triangle, so AB = 1 , or "33. P "3 1 The perimeter is 6(2AB) = 12"33 = 4 " 3. The area is A 1 1 2 ap = 2 (1)(4"3) = B 2 " 3. All rights reserved. 3 Student gives clear and correct diagrams and explanations. 2 Student gives diagrams and explanations that may contain some minor errors. 1 Student gives diagrams or explanations that contain major errors or omissions. 0 Student makes little or no effort. TASK 4:Scoring Guide Sample: a. a circle b. a and r are nearly equal; p and C are nearly equal. c. If the number of sides is great enough that the polygon becomes indistinguishable from a circle, the two formulas are equivalent, and one can be derived from the other. For example: A = 12 ap A = 12 rC A = 12r(2pr) (r a and p C) (C = 2pr) = pr2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3 Student gives correct answers to parts a and b and gives an adequate explanation relating the two formulas in part c. 2 Student gives correct answers to parts a and b but exhibits difficulty with the logic ofpart c. 1 Student gives answers and an explanation that may contain significant errors. 0 Student makes little or no effort. Cumulative Review 1. B 8. H 14. J 2. G 3. D 4. J 5. C 6. J 7. C 9. A 10. G 11. B 12. J 13. A 15. A 16. J 17. Sample: You may have used the branches to climb into the tree. 18. Sample:Suppose that an equilateral triangle has an obtuse angle. Then one angle in the triangle has measure greater than 90. This contradicts the fact that, in an equilateral triangle, the measure of each angle is 60. 19. Sample: 8 cm 5 cm 2 cm 5 cm 20. 16p - 24 " 3 42 21. 61.5 Answers m2 22. Euclid GeometryChapter 10 Chapter 11 Answers Practice 11-1 1. 8 2. 5 3. 12 4. C 5. D 6. B 8. triangle 9. rectangle 10. rectangle 11. rectangle Reteaching 11-1 7. A 1. Sample: 5 in. 8 in. 8 in. 3 in. 4 in. 5 in. 12. rectangle 3 in. 4 in. 4 in. 4 in. 4 in. 3 in. 5 in. 3 in. 3 in. 8 in. 3 in. 5 in. All rights reserved. Practice 11-2 1. 62.8cm2 2. 552.9 ft2 3. 226.2 cm2 4. 113.1 m2 2 2 2 5. 44.0 ft 6. 84.8 cm 7a. 320 m 7b. 440 m2 8a. 576 in.2 8b. 684 in.2 9a. 216 mm2 9b. 264 mm2 10a. 48 cm2 10b. 60 cm2 2 2 11a. 1500 m 11b. 1800 m 12a. 2000 ft2 2 2 12b. 2120 ft 13. 8p m 14. 94.5p cm2 2 15. 290p ft 3 in. 3 in. 4 in. 5 in. 5 in. 2. Checkstudents’ work. 3. vertices = 6; faces = 5; ? edges = 9 4. E = V + F - 2; 9 = 6 + 5 - 2; 9 = 9 5. 6. 7. 8. Practice 11-3 1. 64 m2 2. 620 cm2 3. 188 ft2 4. 36p cm2 2 2 5. 800p m 6. 300p in. 7. 109.6 m2 8. 160.0 cm2 2 2 2 9. 387.7 ft 10. 147.6 m 11. 118.4 cm 12. 668.2 ft2 © Pearson Education, Inc., publishingas Pearson Prentice Hall. Practice 11-4 1. 1131.0 m3 2. 94,247.8 cm3 3. 7.1 in.3 3 3 4. 1781.3 in. 5. 785.4 cm 6. 603.2 cm3 3 3 7. 120 in. 8. 35 ft 9. 42 m3 10. 1728 ft3 3 3 11. 784 cm 12. 265 in. 13. 76 ft3 14. 943 in.3 3 15. 1152 m 1.–2. Check students’ work. 3. L.A. = 300 cm2; S.A. = 350 cm2 4. 722.6cm2 5. 864 in.2 6. 153.3 m2 Practice 11-5 Reteaching 11-3 1. 34,992 cm3 2. 400 in.3 3. 10,240 in.3 3 3 4. 4800 yd 5. 150 m 6. 48 cm3 7. 628.3 cm3 8. 314.2 in.3 9. 4955.3 m3 10. 1415.8 in.3 11. 1005.3 m3 12. 18.8 ft3 13. 20 14. 2 15. 6 1. Sample: Reteaching 11-2 Practice 11-6 1. 2463.0 in.2 2.6,157,521.6 m2 3. 12.6 cm2 2 2 4. 1256.6 m 5. 50.3 ft 6. 153.9 m2 7. 1436.8 mi3 3 3 8. 268,082.6 cm 9. 7238.2 m 10. 14.1 cm3 3 11. 2,572,354.6 cm 12. 904,810.7 yd3 13. 546 ft2 2 2 3 14. 399 m 15. 1318 cm 16. 233 cm 17. 7124 cm3 18. 5547 cm3 19. 42 cm3 Practice 11-7 1. 7 : 9 2. 5 : 8 3. yes; 4 : 7 4.yes; 4 : 3 5. not similar 6. yes; 4 : 5 7. 125 cm3 8. 256 in.3 3 2 2 9. 432 ft 10. 25 ft 11. 90 m 12. 600 cm2 13. 54,000 lb 14. 16 lb Geometry Chapter 11 2. Check students’ work. 3. Sample: base = 3 cm, slant height = 4 cm 4. Sample: 33 cm2 5. 896 cm2 6. 314.2 m2 7. 416.2 in.2 8. 349.3 ft2 Reteaching 11-4 1. 490 cm3 5. 756 ft3 2. 242.5 cm3 6. 2261.9 cm3 3. 735 m3 4. 2544.7 in.3 Reteaching 11-5 1. B 2. 301.6 cm3 5. 211.6 m3 3. 1296 ft3 4. 48 in.3 Answers 35 Chapter 11 Answers (continued) 6. 25 cubic units S.A. = 221.7 mm2 z 1. 2 : 1 2. 3 : 2 3. 64 : 125 4. 512 : 125 5. 250 "2 : 27 6. 16 : 9 7. 49 : 25Enrichment 11-1 1. 8 2. 5 3. 6 4. 11 5. 4 6. 18 7. 7 8. 0 9. 10 10. 15 11. 3 12. 12 13. 2 14. 14 15. 9 16. 25 17. 34 18. 20 19. 22 20. 30 Enrichment 11-2 in.2 1. 96 2. 8 3. 88 4. 4 5. 22 in.2 6. 11 in. 7. 96 in.2 8. (2s2) in.2 9. (96 - 2s2) in.2 10. 4 11. (24 – 0.5s2) in.2 12. s in. 13. ( 24 14. 96 in.2 s - 0.5s) in. 2 15.96 in. ; no matter what the size of the base is, the surface 1. on a line perpendicular to the base at the center of the square 2., 4. 3. isosceles triangle 4. the height of the pyramid 5. Pythagorean Theorem 6. h = 12"4l2 2 s 2 2 8. s6 "4l 2 2 s 2 V 9., 13. M 1. a2 2. b2 3. trapezoid 4. 12 l(a + b) 2 2 5. 2l(a + b)6. a + b + 2l(a + b) 7. πR2 2 2 2 8. pr 9. pRl - prl 10. p(R + r + Rl + rl) 11. 152 cm2 12. 722.6 in.2 13. 506.6 ft2 2 14. 1952 m Enrichment 11-4 1. 48 cubic units 4. 15 cubic units 5. 90 cubic units 2. 30 cubic units z (3, 5, 6) (0, 0, 6) (0, 5, 6) (3, 0, 6) (3, 0, 0) (0, 5, 0) y (3, 5, 0) (0, 0, 0) 7. s2 h a C 10. slantheight 11. apothem 13. h = "l 2 2 a2 14. A = 12. height 2 2 15. nas 2 "l 2 a nas 2 Enrichment 11-6 1. 43 pr 3 2. 34 pR3 3. 43 p(r 3 - R3) 3 4. 266,000,000,000 mi 5. 34 p(r - R)(r 2 + rR + R2) 4 6. 3 pd(3R3) = 4pdR3 7. R d Exact Volume of Shell Approximate Volume of Shell 10 0.1 126.93 125.66 50 132,048.44 31,415.93 100 3 388,413.95 376,991.12 8. 0.421 in.3 9. 1.34 in.3 Enrichment 11-7 1. 180 ft; 120 ft 2. $118.13 3. 6.53 in. 4. 5.52 in. 5. $5.40 6. 8.14 in. 7. 13.58 in. 8. 16.98 in. 9. 3.43 in. 10. 19.48 in. x 36 3. 64 cubic units h s 2 area of the rectangular prism will always be 96 in.2 as long as theentire piece of paper is used. Enrichment 11-3 y Enrichment 11-5 LEONHARD EULER WAS A FAMOUS MATHEMATICIAN KNOWN FOR HIS CONTRIBUTIONS TO GEOMETRY, CALCULUS, AND NUMERICAL ANALYSIS. in.2 (0, 0, 0) (5, 0, 0) (5, 0, 2) x Reteaching 11-7 in.2 (0, 0, 2) (0, 5, 2) (0, 5, 0)Answers All rights reserved. 1.–4. Check students’ work. 5. four circles 6. They are the same. 7. V = 113.1 in.3; S.A. = 113.1 in.2 8. V = 523.6 cm3; S.A. = 314.2 cm2 9. V = 7238.2 m3; 2 3 S.A. = 1809.6 m 10. V = 8181.2 ft ; S.A. = 1963.5 ft2 3 11. V = 1047.4 in. ; S.A. = 498.8 in.2 12. V = 310.3 mm3;Geometry Chapter 11 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Reteaching 11-6 Chapter 11 Answers (continued) ✔ Checkpoint Quiz 2 Chapter Project 1. 201.1 m2 ; 268.1 m3 2. 452.4 ft2 ; 402.1 ft3 3. 158.7 in.2 ; 84 in.3 4. 534.1 yd2 ; 942.5 yd3 5. 384 m2 ; 384 m3 6. 113.1 cm2 ;50.3 cm3 Activity 1: Measuring Check students’ work. Activity 2: Analyzing Sample: l w d V S.A V : S.A. 6 6 6 216 216 1:1 2 3 36 216 372 18 : 31 3 3 24 216 306 36 : 51 4 6 9 216 228 18 : 19 7. 21.2 in.3 Chapter Test, Form A 1. Sample: 2. Sample: 6 ft 6 ft All rights reserved. 3m 1. 6 by 6 by 6; 2 by 3 by36; the lower the ratio, the more material is being used to enclose the same volume. 2. to minimize the cost of packaging 3. Sample: A cereal box is used for promoting the product, so it needs a large surface area on the front. Activity 3: Investigating Check students’ work. Activity 4: Creating Checkstudents’ work. ✔ Checkpoint Quiz 1 1. 6 in. 4 in. 4 in. 4 in. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2 in. 4 in. 3. The lateral area of a paint roller determines the amount of paint it can spread on a wall in one revolution. Because the lateral area of roller A is 24p in.2 and the lateralarea of roller B is 24p in.2, they both spread the same amount of paint in one revolution. 4. V = 400.0 cm3; S.A. = 360.0 cm2 3 5. V = 254.5 in. ; S.A. = 226.2 in.2 6. V = 1050.0 cm3; 2 3 S.A. = 734.0 cm 7. V = 1005.3 ft ; S.A. = 628.3 ft2 3 8. V = 113.1 m ; S.A. = 113.1 m2 9. V = 119 cm3; 2 S.A. = 174 cm10a. a cylinder with radius 2 units and height 3 units 10b. 12p cubic units 10c. 12p square units 11a. half of a sphere and a cone 11b. 251.3 cm2 12. Two regular cans of soup each have a volume of 423.3 cm3 for a total of 846.7 cm3. The single family-size can with its volume of 863.9 cm3 is larger thanthe total volume of the two regular cans. 13. 729 : 64 14. 9 : 4 15. 137.2p cm3 16. S.A. = 603.2 m2; V = 1206.4 m3 17. S.A. = 3327.7 cm2; V = 12,250.6 cm3 18. 18 cm Chapter Test, Form B 2 in. 2. 10 m 7 ft 1. Sample: 4m 2. Sample: 2 cm 10 cm 6m 22 in. 4m 4 cm 3. 88 in.2 4. 80p m2 6. 8. rectangleGeometry Chapter 11 8 in. 5. 1814.4p 7. in.2 2 cm 3. The lateral area of a paint roller determines the amount of paint it can spread on a wall in one revolution. Because the lateral area of roller A is 22p in.2 and the lateral area of roller B is 21p in.2, roller A spreads more paint on a wall in one revolution. 4.V = 523.6 m3; S.A. = 314.2 m2 5. V = 1583.4 cm3; S.A. = 754.0 cm2 6. V = 67.0 ft3; S.A. = 138.2 ft2 7. V = 880.0 in.3; S.A. = 696.0 in.2 8a. half of a sphere and a cone 8b. 301.6 cm2 9. Two regular cans of soup each have a volume of 461.8 cm3 for a total of 923.6 cm3. The single family-size can with itsvolume of 942.5 cm3 is larger than the total volume of the two regular cans. 10. D 11. S.A. = 1357.2 m2; V = 4071.5 m3 12. S.A. = 3296.8 cm2; V = 12,113.3 cm3 Answers 37 Chapter 11 Answers (continued) Alternative Assessment, Form C TASK 3: Scoring Guide TASK 1: Scoring Guide a. Sample:Sample: 2 cm 4 8 8 4 cm 4 3 cm 4 8 4 cm Prism: S.A. = 60 cm2; V = 24 cm3 Cylinder: S.A. = 10p cm2 or about 31.4 cm2; V = 4p cm3 or about 12.6 cm3 3 Student gives accurate drawings and calculations. 2 Student gives drawings and calculations that may contain minor errors. 1 Student gives drawingsand calculations that contain significant errors or omissions. 0 Student makes little or no attempt. 3 Student draws an accurate net and gives correct answers. 2 Student draws a net and gives answers that may contain minor errors. 1 Student draws a net and does calculations incorrectly. 0 Student makeslittle or no attempt. TASK 4: Scoring Guide Foundation drawing: 1 All rights reserved. b. V 390 m3; S.A. 306 m2 4 cm 1 Sample: Pyramids have polygonal sides, whereas cones have rounded sides. Each of these figures has a vertex, and each has volume 1 3 Bh. The base of a pyramid is a polygon, andthe base of a cone is a circle. s a pyramid with square base, height h, slant height s A 4( 12 xs) 2xs V = 13 x2 h a cone with radius r, height h, and slant height s S.A. V = pr 2 + = 31 pr 2h prs 3 Student gives accurate drawings and formulas. 2 Student gives drawings and formulas that may contain minorerrors. 1 Student gives drawings and formulas that contain significant errors. 0 Student makes little or no attempt. 38 2 Orthographic drawing: Front x S.A. 2 Answers 1 r x = x2 + = x2 + 2 s h h 2 Top Right 3 Student draws completely accurate figures. 2 Student draws generally accurate figures, but thesemay contain a minor error. 1 Student draws figures containing significant errors. 0 Student makes little or no attempt. Cumulative Review 1. B 2. J 3. D 4. J 5. C 6. G 7. D 8. H 9. B 10. G 11. B 12. F 13. A 14. J 15. between 4 yd and 12 yd 16. 2 cylinders 17. 14p in.2 18. Sample: A sharpened pencil is acomposite of a cone and a cylinder. The cone is the top of the pencil with the point, and the trunk of the pencil is the cylinder. 19. Check students’ work. 20. Both shapes are triangles, and the sides are proportional in a 2 : 3 ratio. Geometry Chapter 11 © Pearson Education, Inc., publishing as PearsonPrentice Hall. TASK 2: Scoring Guide Chapter 12 Answers Practice 12-1 1. 32 2. 50 3. 72 4. 15 7. "634 8. "901 9. 4 "3 11. inscribed 12. circumscribed 14. 28 in. 15. 52 ft 5. "91 6. 6 10. circumscribed 13. 24 cm y (0, 5) 17. (0, 0) (5, 0) x 5 Practice 12-2 0 1. r = 13; mAB 134.8 3. r = 0 #41 2 ; mAB 102.7 2. r =3 "5; mAB 53.1 0 4. 3 5. 4.5 7. 20.8 8. 13.7 9. 8.5 10. Q T; 0 0 PR SU 11. A J; BC KL 12. Construct radii OD and OB . FD = EB because a diameter perpendicular to a chord bisects it, and chords CD and AB are congruent (given). Then, by HL, OEB OFD, and by CPCTC, OE = OF. 13. Because congruentarcs have congruent chords, AB = BC = CA. Then, because an equilateral triangle is equiangular, mABC = mBCA = mCAB. All rights reserved. 6. 3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Practice 12-3 1. A and D; B and C 2. ADB and CDB 3. ADB and CAD 4. 55 5. x = 45; y = 50;z = 85 6. x = 90; y = 70 7. 180 8. 70 9. x = 120; y = 60; z = 60 10. x = 120; y = 100; z = 140 11. x = 63; y = 63; z = 540 12. x = 50; y = 80; z = 80 13a. mAE = 170 13b. mC = 85 13c. mBEC = 10 13d. mD = 85 14a. mA = 90 14b. mB = 80 14c. mC = 90 14d. mD = 100 Practice 12-4 y 18. 3 5 4 3 2 1 (3, 5) y19. 2 2 2 x 4 6 20. (1, 7) y 6 4 (1, 1) 4 2 Practice 12-5 Geometry Chapter 12 4 2 (2, 4) 4 1. 87 2. 35 3. 45 4. 120 5. 72 6. 186 7. x = 58; y = 59; z = 63 8. x = 30; y = 66 9. x = 30; y = 30; z = 120 10. x = 16; y = 52 11. x = 138; y = 111; z = 111 12. x = 30; y = 60 13. 10 14. 14.8 15. 4.7 16. 4 17. 3.2 18. 6 1.C(0, 0), r = 6 2. C(2, 7), r = 7 3. C(-1, -6), r = 4 4. C(-3, 11), r = 2"3 5. x2 + y2 = 49 6. (x - 4)2 + (y - 3)2 = 64 7. (x - 5)2 + (y - 3)2 = 4 8. (x + 5)2 + (y - 4)2 = 14 9. (x + 2)2 + (y + 5)2 = 2 10. (x + 1)2 + (y - 6)2 = 5 11. x2 + y2 = 4 2 2 12. (x + 3) + (y - 3) = 1 13. x2 + (y - 3)2 = 16 2 2 14. (x - 7) + (y + 2) = 4 15. x2+ (y + 20)2 = 100 2 2 16. (x + 4) + (y + 6) = 25 x 1 2 3 4 5 6 2 (5, 1) 2 4 x 2 4 21. 23. 24. 25. 26. x2 (x (x (x (x + + + - y2 = 25 4)2 + (y 7)2 + (y 4)2 + (y 4)2 + (y 22. (x - 5)2 + (y - 9)2 = 9 + + + - 3)2 2)2 3)2 1)2 = = = = 61 80 4 16 Answers 33 Chapter 12 Answers (continued) 11. Practice 12-6 1. 2 cm Q T 1.5cm 12. 0.75 in. 2. 1 in. Q S 0.75 in. T U 13. 0.5 cm 3. 14. A B 5 mm 6 mm 4. All rights reserved. P R 0.75 in. 0.5 in. R 0.5 in. Reteaching 12-1 S 0.75 in. 1. "157 2. "741 3. "2 4. 7.5 Reteaching 12-2 5. B A E C 7. H F Reteaching 12-3 1. 87 6. 120 2. 40 7. 40 3. 60 8. 20 4. 55 9. 70 5. x = 94, y = 80Reteaching 12-4 1. 93 2. 156 3. 42 7. x = 36; y = 60; z = 48 9. x = 46; y = 90; z = 44 4. 35 5. 60 6. 55 8. x = 64; y = 64; z = 52 Reteaching 12-5 P J G 8. a third line parallel to the two given lines and midway between them 9. a sphere whose center is the given point and whose radius is the given distance10. the points within, but not on, a circle of radius 1 in. centered at the given point 1. (x - 3)2 + (y 3. (x - 6)2 + (y + 5. (x + 2)2 + (y + 6. (x - 3)2 + (y 7. (x - 5)2 + (y 8. (x - 4)2 + (y 9. (x + 2)2 + (y 11. C(0, 0); r = 0.2 13. C(3, 5); r = 4 11)2 = 4 2. (x + 5)2 + y2 = 225 2 6) = 7 4. x2 + y2 = 20 2 2) = 4 1)2 = 50 2)2 =36 2)2 = 25 3)2 = 25 10. C(-3, -5); r = 5 12. C(4, 0); r = "6 Reteaching 12-6 1.–5. Check students’ work. 6. the line perpendicular to AB at its midpoint 7. one, the midpoint of AB 34 Answers Geometry Chapter 12 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 6. 1. 5 "5 2. 12 "2 3. 2 "14 4.11.83 cm 5. 2.54 cm 6. 8.66 in. 7. 2.78 in. Y X All rights reserved. Chapter 12 Answers (continued) Enrichment 12-1 Enrichment 12-6 1. Given 2. Two points determine a line segment. 3. Two tangents drawn to a circle from an external point are congruent. 4. Radii of a circle are congruent. 5. A radius anda tangent drawn to the same point of contact form a right angle. 6. Definition of a square 7. Sides of a square are congruent. 8. Addition Property 9. Segment Addition Postulate 10. Substitution Property 11. Addition Property 12. Substitution Property 13. Subtraction Property 1.–9. Check students’drawings. 1. P with a radius of 4 in. 2. two lines parallel to line l, one 2 in. above l and the other 2 in. below l 3. 4 points 4. two lines 5. 2 points Enrichment 12-2 Activity 1: Doing 1a. 1 1b. 3 1c. 6 1d. 10 1e. 15 N(N 2 1) 2. 1, 3, 6, 10, 15 3. triangular numbers 4. 2 5. From each point, (N - 1) chords may bedrawn. So, a total Activity 2: Exploring of N(N - 1) chords may be drawn for N points. Because each chord has been drawn twice, divide by 2. 6. N(N - 1) Enrichment 12-3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. Given 2. Two points determine a line segment. 3. The measures ofexterior angles of a triangle equal the sum of the measures of the remote interior angles. 4. Radii of a circle are congruent. 5. Definition of isosceles triangle 6. Base angles of an isosceles triangle are congruent. 7. Definition of congruent angles 8. Base angles of an isosceles triangle are congruent. 9.Definition of congruent angles 10. The measures of exterior angles of a triangle equal the sum of the measures of the remote interior angles. 11. Substitution Property 12. Division Property 13. drawing diameter DE passing through A so that radius OB AB Enrichment 12-4 1. 2x + 2x + 8 + x + x - 32 = 360;6x - 24 = 360; 6x = 384 2. x = 64 3. 128 4. 136 5. 64 6. 32 7. chords AB , BD ; 16 8. secants EB , EC; 52 ) 9. tangent FB and chord AB; 64 10. chords BD , AC; 84 ) 11. chords AC, BC; 64 12. tangent FB and secant FD; 36 13. chords AC, BD; 96 14.) AF, AE; 100 15. DE, DA; 48 16. tangent FB and chordBC; 68 Enrichment 12-5 parallel to line l, one 4 in. above l and the other 4 in. below l 6. two lines parallel to line l, one 6 in. above l and the other 6 in. below l 7. empty set 8. 6 points 9. empty set Chapter Project Check students’ work. Both figures are made from 9 circles with the same center. In Figure A,the smallest circle and alternate rings are black. In Figure B, vertical and horizontal tangents to the smallest circle are drawn. Then the tangents at 45° angles to the first ones are drawn. Alternating sections of the 6 outer rings are black, as is the smallest ring. Check students’ work. Activity 3: ConstructingCheck students’ work. ✔ Checkpoint Quiz 1 1. 52 in. 6. 11.5 z = 79 Geometry Chapter 12 4. 6 5. 12 8. x = 116, y = 88, ✔ Checkpoint Quiz 2 1. x = 140 2. x = 75, y = 105 3. x = 20, y = 70 4. x2 + (y - 1)2 = 6.76 5. (x + 3)2 + (y - 2)2 = 100 6. 12 7. 51 8. 20 7 Chapter Test, Form A 1. 3. 5. 7. 9. C(0, 0); r = 10 2.C(11, -6); r = 9 C(-1, -4); r = "7 4. (x + 3)2 + (y - 2)2 = 1 2 2 (x - 2) + (y - 1) = 16 6. x2 + (y - 2)2 = 9 C = 31.4; A = 78.5 8. (x + 3)2 + (y - 4)2 = 41 y C(1, 2) r2 1a. the edge of the fountain pool 1b. plaza 1c. fountain pool 2. a circle with center (-3, 4) and a radius of 3 3. all the interior points of a circle with center(4, 0) and a radius of "12 4. all points outside a circle with center (0, 0) and a radius of "18 5. a circle and all its interior points with center (0, -2) and a radius of 5 6. x 2 + y 2 = 100 7. x2 + y2 36 8. The sergeant can draw a circle with center (0, 0) and a radius of 2. The team will search in the circle. 2. 44 cm3. 40 m 7. x = 37, y = 100 9. x = 120 x 10. 90 Answers 35 Chapter 12 Answers 11. (continued) y Chapter Test, Form B x 0 0 1. inscribed 2. circumscribed 3. R; QS 4. M; LN 5. center (0, 0); r = 5 6. center (3, 5); r = 10 7. x2 + y2 = 9 8. (x - 4)2 + (y + 1)2 = 4 9. 37.7 units; 113.1 square units 10. x2 + (y + 1)2

= 25 11. center (-3, 2); r = 3 y 6 4 2 y 6 4 2 2 2 4 x 6 4 6 x 12. 90 13. 106 14. 105 15. 90 16. 45 17. 90 18. 180 19. 270 20. 47 21. 10 22. 8 23. 22 24. 126 25. 55 All rights reserved. 12. Alternative Assessment, Form C x 5 TASK 1: Scoring Guide y (a) center (2, 5), radius = 4 (b) The distance from point (x,y) on a circle to the center (2, 5) is given by the formula D = "(x 2 2) 2 1 (y 2 5) 2. In any circle, the distance, or radius, is the same for all points on the circle. Therefore the distance formula, when applied to the circle in this problem, becomes 4 = "(x 2 2) 2 1 (y 2 5) 2 for all points (x, y) on the circle.Squaring both sides of this equation yields the equation of the circle, 16 = (x - 2)2 + (y - 5)2. x 14. y y4 3 Student gives accurate answers and a correct explanation. 2 Student gives answers or an explanation that may contain minor errors. 1 Student gives wrong answers or an incomplete or inaccurateexplanation. 0 Student makes little or no effort. x TASK 2: Scoring Guide y 6 15. 120 20. 115 25. 35 16. 105 17. 118.0 18. 40 19. 70 21. 6 22. 11.8 23. 5.5 24. 45 26. 80 27. 5.8 28. 5.7 29. x = 63; y = 71 30. x = 87; y = 95; w = 85; z = 93 31. x = 120; y = 64; z = 176 32. 44 36 Answers (a) Construct theperpendicular bisector of two chords. The point at which they meet is the center of the circle. (b) Because the pentagon is regular, all chords are congruent. Therefore the corresponding arcs are congruent. Because there are 50 congruent arcs in the circle, each arc, and in particular AB , has measure 72.To find AB, call the center O, 0 and consider triangle AOB. Because AB has measure 72, so does angle AOB. Because OA = OB, AOB is isosceles. Therefore, the bisector of angle AOB is perpendicular to (and bisects) chord AB, forming a 36°-54°-90° triangle. Then 1 AB AB , yielding AB = 7.05. = sin 36= 2 12 OA Geometry Chapter 12 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 13. Chapter 12 Answers (continued) 3 Student devises a correct method and gives a correct answer and a valid explanation. 2 Student devises a method, gives an answer, and gives an explanation that maycontain some errors. 1 Student gives a method, an answer, and an explanation that may contain major errors or omissions. 0 Student makes little or no effort. All rights reserved. TASK 3: Scoring Guide (a) Angles A and C are inscribed angles. Each is inscribed in a different arc, but together the arc inwhich they are inscribed comprise the entire circle. Therefore, mA + mC = 12(360) = 180. And, because ABCD is a parallelogram, angles A and C are congruent. Finally, if angles are congruent and supplementary, then they are right. (b) The diagonals of ABCD bisect each other because ABCD is aparallelogram. Suppose that they intersect at point X. Then AX = CX and BX = DX. Also, because ABCD is inscribed in a circle, the diagonals are chords, and therefore AX ? CX = BX ? DX. Substituting yields AX2 = BX2, and therefore AX = BX = CX = DX. Then AC = BD. (c) Either (a) or (b) allows you toconclude that ABCD must be a rectangle. Cumulative Review 1. B 8. F 14. H 2. J 3. A 4. F 5. D 6. G 7. A 9. D 10. G 11. C 12. H 13. D 15. B 16. "89 17. Check students’ work. 18. A tangent intersects a circle at exactly one point. A secant intersects a circle at two points. 19. Check students’ work. 20. Thisfollows from the Pythagorean Theorem (c2 = a2 + b2) or the fact that the longest side of a triangle is opposite the largest angle. 21. If a triangle is not a right triangle, then it is acute. 22. 135 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3 Student gives valid and accurate arguments andexplanations. 2 Student gives arguments that, although basically valid, may contain minor flaws. 1 Student gives arguments containing major flaws. 0 Student makes little or no attempt. TASK 4: Scoring Guide (a) Because the circumscribing lines are tangent to the circle, AB = AX, CB = CY, and DX = DY.Then, because it is given that AB = BC, we have AX = CY. Adding segments together yields DA0 = DC, showing that1 ACD is isosceles. Secondly, because XY has measure 100, XBY has measure 260. Then mD = 21 (260 - 100) = 80, and because ACD is isosceles, mA = mC = 50. Finally, AC = 20 ABbecause AB = BC = 10, and, by trigonometry, cos 50 = AD , 10 AB AD = cos 50 = 0.6428 = 15.56. And again, AD = CD. (b) Because angle A has measure 50 and line AB is tangent at B, triangle AOB is a 25°-65°-90° triangle. Therefore, by OB trigonometry, tan 25 = AB , OB = 10 ? tan 25 = 4.66. 3 Studentgives accurate answers and explanations. 2 Student gives answers and explanations that may contain minor errors. 1 Student gives answers and explanations that contain significant errors. 0 Student makes little or no attempt. Geometry Chapter 12 Answers 37 Geometry: All-In-One Answers Version A 2Practice 1-1 Front 6. Right 1 Guided Problem Solving 1-1 1 1. The pattern is easier to visualize. 2. The graph will go up. 3. Use Years for the horizontal axis. 4. Use Number of Stations for the vertical axis. 5. increasing 6. greater 7. Yes. Since the number of stations increases steadily from 3 1950 to 2000,we can be confident that the number of stations in 2010 will be greater than in 2000. 8. Patterns are necessary to reach a conclusion through inductive reasoning. 9. (any list of numbers without a pattern would apply) 2,435; 16,439; 16,454; 3,765; 210,564 Top Right ratio gets to a number that isapproximately 0.618. 20. 0, 1, 1, 2, 3, 5, 8, 13 1 Front 1. 47, 53 2. 1.00001, 1.00001 3. 42, 54 4. -64, 128 5. 22, 29 6. 63.5, 63.75 7. Sample: 2 or 3 8. 51 or 49 9. 6 or 8 10. A or AA 11. D or G 12. Y or A 13. any hexagon 14. circle with 8 equally spaced diameters 15. a 168.75° angle 16. 21 hand-shakes 17.h = n(n 22 1) 18. 34 19. Sample: The farther out you go, the closer the All rights reserved. 2 Right 5. Chapter 1 Front Front Top Right 7. A, C, D 8. C 9. D 10. B 11. A Guided Problem Solving 1-2 1. They represent three-dimensional objects on a twodimensional surface. 2. nine 3. See the figure in 4, below.4. Practice 1-2 Fro nt Rig ht 2. Fro 5. Yes. It is similar to the foundation drawing, except there are no numbers. 6. no nt Rig 7. ht 3. Fro Top nt Rig 4. 1 1 1 1 1 1 ht Front 8. Yes. 9. Right © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. 4 2 3 1 Right Front Practice * ) 1-3 Front Top All-In-One Answers Version A Right 1. AC 2. any two of the following: ABD, DBC, CBE, ABE, ECD, ADE, ACE, ACD 3. Points E, B, and D are collinear. 4. yes 5. yes 6. no 7. no 8. yes 9. no Geometry 63 Geometry: All-In-One Answers Version A yes 11.* yes) 12. yes * ) 13. no 14. yes 15. yes* ) G 17. LM 18. PN19. the empty set 20. KP *M ) 22. Sample: plane ABD 23. Sample: plane ABC AB 25. yes 26. no 27. yes 28. the empty set no 30. yes 31. yes 5. Answers may vary. Sample: (0, 0) (Answers will be coordinates (x, y), where y = 32x, x < 2.) y 6 Guided Problem Solving 1-3 4 1. Collinear points lie on the sameline. 2. Answers may vary. (Some people might note that the y-coordinate of two of the points is the same so that the third point must have the same y-coordinate to be collinear. Since it does not, the points are not collinear.) 3. horizontal 4. No. 5. No. 6. All points must have the same y-coordinate, -3. 7.No. 8. a1, 21 b 2 A(2,3) 2 x 2 2 2 C(0,0) 4 6 6. yes 7. L(4, 2) 1. true 2. false 3. true 4. false 5. false 6. false * ) 7. JK, HG 8. EH 9. any three of the following pairs: EF * )* ) * )* ) * )* ) and ) *and JH ) * ; )EF and * )GK * ;) HG* and) JE * ; )HG *and) FK * ); JK *KG EH ; JK and FG ; EJ and FG ; EH and FK ; JEand ; * ) * ) *KG ) *JH ) * ) *JH ) GE EH and ; and KF ; and 10. planes A and B 11. planes A and C; planes B and C 12. planes A and C ) ) 13. planes B and C 14. Sample: EG 15. 6 16. EF and ) ) ) ) ) ) ) ED or EG and ED 17. FE , FD 18. GF , GD 19. yes 20. Sample: Practice 1-5 1. 4 2. 12 3. 20 4. 6 5. 22 6.-10 or 6 7. -1 or 1 8. 3; 4; no 9. 6; 6; yes 10. C or -2 11. 15 12. 31 13. 14 14. x = 11 32 ; AB = 31; BC = 31 15. x = 35 23 ; AB = 103; BC = 103 Guided Problem Solving 1-5 T 1. AD DC 2. AD = DC 3. Segment Addition Postulate. 4. Since AD = DC, AC = 2(AD). 5. AC = 2(12) = 24 6. y = 15 7. DC = AD = 12 8.Answers may vary. 9. ED = 11, DB = 11, EB = 22 U Practice 1-6 S 1. any three of the following: &O, &MOP, &POM, &1 2. &AOB 3. &EOC 4. &DOC 5. 51 6. 90 7. 17 8. 107 9. 141 10. 68 11. &ABD, &DBE, &EBF, &DBF, &FBC 12. &ABF, &DBC 13. &ABE, &EBC 14. x = 5; 8; 21; 13 15. x = 9; 85; 35; 12021. Sample: R Q W P Guided Problem Solving 1-6 1. Angle Addition Postulate 2. supplementary angles 3. m&RQS + m&TQS = 180 4. (2x + 4) + (6x + 20) = 180 5. x = 19.5 6. m&RQS = 43; m&TQS = 137 7. The sum of the angle measures should be 180; m&RQS + m&TQS = 43 + 137 = 180. 8a. x = 118b. m&AOB = 17; m&COB = 73 Guided Problem Solving 1-4 1. Opposite rays are two collinear rays with the same endpoint. 2. a line 3–4. See graph in Exercise 5 answer. Practice 1-7 1. X 64 Geometry All rights reserved. Practice 1-4 * )* ) B(4,6) Y A B All-In-One Answers Version A © PearsonEducation, Inc., publishing as Pearson Prentice Hall. 10. 16. 21. 24. 29. (continued) Geometry: All-In-One Answers Version A (continued) 2. 9. 1 2 1 X 2 B Y 10. 2 ? 2 2 3. C All rights reserved. 11. X Y X X 4. 12. 2 ? X Z W © Pearson Education, Inc., publishing as Pearson Prentice Hall. 5. 13. X 45° Z 90°14. true 15. false 16. false 17. true 18. true 19. false 20. true 6. MN OP Guided Problem Solving 1-7 B A 7. 1. &DBC > &ABC 2. complementary angles 3. &CBD 4. m&CBD = m&CBA = 41 5. m&ABD = m&CBA + m&CBD = 41 + 41 = 82 6. m&ABE + m&CBA = 90 OP MN K L m&ABE + 41 = 90 m&ABE =49 7. m&DBF = m&ABE = 49 8. Answers may vary. Sample: The sum of the measures of the complementary angles should be 90 and the sum of the measures of the supplementary angles should be 180. 9. m&CBD= 21, m&FBD = 69, m&CBA =21, and m&EBA = 69 8. 2 1 A 1 All-In-One Answers VersionA 2 Geometry 65 Geometry: All-In-One Answers Version A Practice 1-8 Guided Problem Solving 1-9 1.–5. 1. six 2. It is a two-dimensional pattern you can fold to form a three-dimensional object. 3. rectangles y 6 A C 4. 4 6 4 2 6 D 4 (continued) 2 E 2 4 4 4 6 6 x 4 8 2 8 B 4 4 6 Q 4 2 P R 2 4 6 8x 4 4 6 65. 208 in.2 6. They are equal. 7. 208 in.2 8. Answers will vary. Sample: 2(4 6) + 2(4 8) + 2(6 8); the results are the same, 208 in.2 9. 6(72) = 294 in.2 1A: Graphic Organizer 1. Tools of Geometry 2. Answers may vary. Sample: patterns and inductive reasoning; measuring segments and angles; basicconstructions; and the coordinate plane 3. Check students’ work. 1B: Reading Comprehension * ) * )* ) * ) 1. Answer may vary. Sample: AB 6 CD , EF 6 GH , JK > LM, JL > KM, m/AJF + m/FJK = 180°, ) * ) /HKB > /KMD,DN ' CD 2. Points A, M, and S are * ) * ) * ) collinear. 3. AB , HI , and LN intersect atpoint M. 4. a 24. 24.7 25. (3.5, 3) 1C: Reading/Writing Math Symbols Guided Problem Solving 1-8 1. Distance Formula 2. The distance d between two points A(x1,y1) to B(x2,y2) is d 5 Í (x2 2 x1) 2 1 (y 2 2 y1) 2. 3. No; the differences are opposites but the squares of the 1. Line BC is parallel to line MN. 2.Line CD 3. Line segment GH 4. Ray AB 5. The length of segment XY is greater than the length of segment ST. 6. MN = XY 7. GH = 2(KL) 8. ST ' UV 9. plane ABC || plane XYZ 10. AB 6 DE differences are the same. 4. XY 5 Í (5 2 (26)) 2 1 (22 2 9) 2 5. To the nearest tenth, XY = 15.6 units. 6. To thenearest tenth, XZ = 12.0 units. 7. Z is closer to X. 8. The results are the same; e.g., XY 5 Í (26 2 5) 2 1 (9 2 (22)) 2 5 Í 242 , or about 15.6 units, as before. 9. YZ 5 Í (17 2 (26)) 2 1 (23 2 9) 2 5 Í 673 ; to the nearest tenth, YZ = 25.9 units. To the nearest tenth, XY+YZ+XZ = 53.5 units. Practice 1-9 1. 792 in.22. 3240 in.2 3. 2.4 m2 4. 32p 5. 16p 6. 7.8p 7. 26 cm; 42 cm2 8. 46 cm; 42 cm2 9. 29 in.; 42 in.2 10. 40 ft; 51 ft2 11. 40 m; 99 m2 12. 40 m; 91 m2 13. 68; 285 14. 26;3 22 15. 30; 44 16. 156.25p 17. 10,000p 18. p16 19. 48; 28 20. 36 21. 26; 13 66 Geometry All rights reserved. 6. 5"2 7.1 7. 2"17 8.2 8. 12 9.8 10. 12 11. "26 5.1 12. (5, 5) 13. (21 , 1) 14. (10 21 , -5) 15. (-2 12 , 6) 16. (-0.3, 3.4) 17. (2 78 , -58 ) 18. (5, -2) 19. (4, 10) 20. (-3, 4) 21. yes; AB = BC = CD = DA = 6 22. "401 20.025 23. y S 4 1D: Visual Vocabulary Practice 1. parallel planes 2. Segment Addition Postulate 3. supplementary angles 4.opposite rays 5. isometric drawing 6. perpendicular lines 7. foundation drawing 8. right angle 9. congruent sides 1E: Vocabulary Check Net: A two-dimensional pattern that you can fold to form a three-dimensional figure. Conjecture: A conclusion reached using inductive reasoning. Collinear points: Pointsthat lie on the same line. Midpoint: A point that divides a line segment into two congruent segments. Postulate: An accepted statement of fact. All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 6 Geometry: All-In-One Answers Version A (continued) All rightsreserved. 1F: Vocabulary Review Puzzle C O L L I N E A R P P D C L Z J P S Y J E Z C C C E O R H L O E G E L K C A F R T S V I L A B B N L I L A E A O R E U N N X W L T Z S L B G N W I N N Y L T I Y Y U R G T A C N E L F Q G G D P C O S C S B R R K A C I E N I L R D M E P I D O U A X TF N R V P V E U A A J D V K C P Y H U E D I C V D N E H X N I I T X S G M S L X S E G M E N T E O M I A V I X K F L P L S D K V T P R C O F J R N E B L E V D V J S S R R E N L W F Q H G N I N O S A E R E V I T C U D N I R O T C E S I B L P J A E N R U K R S T R A I G H T A N G L E W U QA P E L G N A E T U C A W K M F B O Z N T U B A R N K I W T K Q R U I V C F Chapter 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Practice 2-1 1. Sample: It is 12:00 noon on a rainy day. 2. Sample: The car will not start because of a dead battery. 3. Sample: 6 4. If you arestrong, then you drink milk. 5. If a rectangle is a square, then it has four sides the same length. 6. If you are tired, then you did not sleep. 7. If x = 26, then x - 4 = 22; true. 8. If x 0, then ∆x∆ 0; true. 9. If m is positive, then m2 is positive; true. 10. If 2y - 1 = 5, then y = 3; true. 11. If x 0, then point A is in thefirst quadrant; false; (4, -5) is in the fourth quadrant. 12. If lines are parallel, then their slopes are equal; true. 13. If you have a sibling, then you are a twin; false; a brother or sister born on a different day. 14. siblings twin 15. Hypothesis: If you like to shop; conclusion: Visit Pigeon Forge outlets inTennessee. 16. Pigeon Forge outlets are good for shopping. 17. If you visit Pigeon Forge outlets, then you like to shop. 18. It is not necessarily true. People may go to Pigeon Forge outlets because the people they are with want to go there. 19. Drinking Sustain makes you train harder and run faster. 20. Ifyou drink Sustain, then you will train harder and run faster. 21. If you train harder and run faster, then you drink Sustain. Guided Problem Solving 2-1 1. Hypothesis: x is an integer divisible by 3. 2. Conclusion: x2 is an integer divisible by 3. 3. Yes, it is true. Since 3 is a factor All-In-One Answers Version Aof x, it must be a factor of x ? x = x2. 4. If x2 is an integer divisible by 3 then x is an integer divisible by 3. 5. The converse is false. Counterexamples may vary. Let x2 = 3. Then x 5 Í 3 , which is not an integer and is not divisible by 3. 6. No. The conditional is true, so there is no such counterexample. 7.No. By definition, a general statement is false if a counterexample can be provided. 8. If 5x + 3 = 23, then x = 4. The original statement and the converse are both true. Practice 2-2 1. Two angles have the same measure if and only if they are congruent. 2. 2x - 5 = 11 if and only if x = 8. 3. The converse, “If∆n∆ = 17, then n = 17,” is not true. 4. A figure has eight sides if and only if it is an octagon. 5. If a whole number is a multiple of 5, then its last digit is either 0 or 5. If a whole number has a last digit of 0 or 5, then it is a multiple of 5. 6. If two lines are perpendicular, then the lines form four right angles. If twolines form four right angles, then the lines are perpendicular. 7. If you live in Texas, then you live in the largest state in the contiguous United States. If you live in the largest state in the contiguous United States, then you live in Texas. 8. Sample: Other vehicles, such as trucks, fit this description. 9.Sample: Other objects, such as spheres, are round. 10. Sample: This is not specific enough; many numbers in a set could fit this description. 11. Sample: Baseball also fits this definition. 12. Sample: Pleasing, smooth, and rigid all are too vague. 13. yes 14. no 15. yes Guided Problem Solving 2-2 1. Agood definition is clearly understood, precise, and reversible. 2. &3 and &4, &5 and &6 3. No Geometry 67 Geometry: All-In-One Answers Version A 5. y B(0,3) pairs: &1 and &3, &1 and &4, &2 and &3, &2 and &4 A(1,3) Practice 2-3 Guided Problem Solving 2-3 1. conditional; hypothesis 2. Yes 3. Beth willgo. 4. Anita, Beth, Aisha, Ramon 5. No; only two students went. 6. Beth, Aisha, Ramon; no—only two went. 7. Aisha, Ramon 8. The answer is reasonable. It is not possible for another pair to go to the concert. 9. Ramon Practice 2-4 1. UT = MN 2. m&QWR = 30 3. SB = MN 4. y = 51 5. JL 6. Given;Addition Property of Equality; Division Property of Equality 7. Given; Addition; Subtraction Property of Equality; Multiplication Property of Equality; Division Property of Equality 8. Substitution 9. Substitution 10. Transitive Property of Equality 11. Symmetric Property of Congruence 12. Transitive Property ofCongruence 13. Definition of Complementary Angles; 90, Substitution; 3x, Simplify; 3x, 84, Subtraction Property of Equality; 28, Division Property of Equality 14. Given; (2x - 4), Substitution; 6x - 12, Distributive Property; -x, Subtraction Property of Equality; 12, Multiplication Property of Equality C(3,1) 6.They are adjacent complementary angles. 7. Answers may vary. C can be any point on the line y 5 21 x, x . 0. 3 Sample: C(3, -1). 8. a right angle 9. Yes; their sum corresponds to the right angle formed by the positive x-axis and the positive y-axis. 10. Answers may vary. D can be any point on the negativex-axis, sample: D(-4, 0) 2A: Graphic Organizer 1. Reasoning and Proof 2. Answers may vary. Sample: conditional statements; writing biconditionals; converses; and using the Law of Detachment 3. Check students’ work. 2B: Reading Comprehension 1. 42 degrees 2. 38 degrees 3. b 2C: Reading/WritingMath Symbols 1. Segment MN is congruent to segment PQ. 2. If p, then q. 3. The length of MN is equal to the length of PQ. 4. Angle XQV is congruent to angle RDC. 5. If q, then p. 6. The measure of angle XQV is equal to the measure of angle RDC. 7. p if and only if q. 8. a S b 9. AB = MN 10. m/XYZ =m/RPS 11. b S a 12. AB > MN 13. a 4 b 14. /XYZ /RPS Guided Problem Solving 2-4 2D: Visual Vocabulary Practice 1. Angle Addition Postulate 2. Substitution Property of 1. Law of Detachment 2. hypothesis 3. Distributive Property 4. Reflexive Property 5. Law of Syllogism 6. biconditional 7. conclusion 8.good definition 9. Symmetric Property Equality; Simplify; Addition Property of Equality; Division Property of Equality 3. 40 4. yes; yes 5. 13; 13 Practice 2-5 1. 30 2. 15 3. 20 4. 6 5. 16 6. 9 7. m&A = 135; m&B = 45 8. m&A = 36; m&B = 144 9. m&A = 75, m&B = 15 10. m&A = 10; m&B = 80 11. m&PMO = 55;m&PMQ = 125; m&QMN = 55 12. m&BOD = m&COE = 90; m&BOC = 2E: Vocabulary Check m&COD = 45; m&AOB = m&DOE = 45 13. m&BWC = m&CWD; m&AWB + m&BWC = 180; m&CWD + m&DWA = 180; m&AWB = m&AWD and its converse; it contains the words “if and only if.” Conclusion: Thepart of an if-then statement that follows then. Conditional: An if-then statement. Truth value: “True” or “false” according to whether the statement is true or false, respectively Hypothesis: The part that follows if in an if-then statement. Biconditional: The combination of a conditional statement Guided ProblemSolving 2-5 1. 90 2. See graph in Exercise 5 answer. 3. on the positive y-axis 4. Answers may vary. B can be any point on the positive y-axis. Sample: B(0, 3). 68 Geometry All rights reserved. 1. &A and &B are complementary. 2. Football practice is canceled for Monday. 3. DEF is a right triangle. 4. If youliked the movie, then you enjoyed yourself. 5. If two lines are not parallel, then they intersect at a point. 6. If you vacation at the beach, then you like Florida. 7. not possible 8. Tamika lives in Nebraska. 9. not possible 10. It is not freezing outside. 11. Shannon lives in the smallest state in the United States.12. On Thursday, the track team warms up by jogging 2 miles. X(4,0) x O(0,0) All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4. They are not supplementary. 5. A linear pair has a common vertex, shares a common side, and is supplementary. 6. yes 7. yes;yes 8. linear pairs: &1 and &2, &3 and &4; not linear (continued) Geometry: All-In-One Answers Version A (continued) 2F: Vocabulary Review Puzzle 1 3 P O S T U L A T 2 V 4 E P R 5 7 C O G O N S All rights reserved. C T D I J C C L P A R C O L I N A A J T N E E U T G C E A L T M N E U I G S R CT I O 11 N T D 12 H Y P L O T H O E A L L E L N E E U T I N R N D 8 A O I O 6 T A 9 N 10 A C O R E S I S S I N T © Pearson Education, Inc., publishing as Pearson Prentice Hall. Chapter 3 Practice 3-1 1. corresponding angles 2. alternate interior angles 3. same-side interior angles 4. alternate interiorangles 5. same-side interior angles 6. corresponding angles 7. 1 and 5, 2 and 6, 3 and 8, 4 and 7 8. 4 and 6, 3 and 5 9. 4 and 5, 3 and 6 10. m1 = 100, alternate interior angles; m2 = 100, corresponding angles or vertical angles 11. m1 = 75, alternate interior angles; m2 = 75, vertical angles orcorresponding angles 12. m1 = 135, corresponding angles; m2 = 135, vertical angles 13. x = 103; 77°, 103° 14. x = 24; 12°, 168° 15. x = 30; 85°, 85° 16a. Alternate Interior Angles Theorem 16b. Vertical angles are congruent. 16c. Transitive Property of Congruence Guided Problem Solving 3-1 1. The topand bottom sides are parallel, and the left and right sides are parallel. 2. The two diagonals are transversals, and also each side of the parallelogram is a transversal for the two sides adjacent to it. 3. Corresponding angles, interior and exterior angles are formed. 4. v, w and x; By the Alternate InteriorAngles Theorem, v = 42, w = 25 and x = 76. 5. Answers may vary. Possible answer: By the Same-Side Interior Angles Theorem, (w + 42) + (y + 76) = 180. Since w = 25, y = 37. (The two y’s are equal by Theorem 3-1.) 6. w = 25, y = 37, v = 42, x = 76; yes 7. v = 42, w = 35, x = 57, y = 46 All-In-OneAnswers Version A Practice 3-2 * ) * ) 1a. same-side interior 1b. QR 1c. TS 1d. same-side * ) interior 1e. Same-Side Interior Angles 1f. TS 1g. 3-5 2. l and m, Converse of Same-Side Interior Angles Theorem 3. none 4. BC and AD , Converse of Same-Side Interior Angles Theorem 5. RT and HU , Converseof Corresponding Angles Postulate 6. BH and CI, Converse of Corresponding Angles Postulate 7. a and b, Converse of Same-Side Interior Angles Theorem 8. 43 9. 90 10. 38 11. 100 12. 70 13. 48 Guided Problem Solving 3-2 1. and m 2. transversals 3. x 4. the angles measuring 19xº and 17xº 5. 180º 6.17xº 7. 180 - 19x = 17x or 19x + 17x = 180 8. x = 5 9. With x = 5, 19x = 95 and 17x = 85. 10. x = 6 Practice 3-3 1. True. Every avenue will be parallel to Founders Avenue, and therefore every avenue will be perpendicular to Center City Boulevard, and therefore every avenue will be perpendicular to anystreet that is parallel to Center City Boulevard. 2. Not necessarily true. No information has been given about the spacing of the streets. 3. True. The fact that one intersection is perpendicular, plus the fact that every street belongs to one of two groups of parallel streets, is enough to guarantee that allintersections are perpendicular. 4. True. Opposite sides of each block must be of the same type (avenue Geometry 69 Geometry: All-In-One Answers Version A l1 1. A theater stage, consisting of a large platform surrounding a smaller platform. The shapes in the bottom part of the figure may represent aramp for actors to enter and exit. 2. The measures of angles 1 and 2 3. 8; octagon 4. (8 - 2)180 = 1080 degrees 5. 135 6. 45 7. Yes, angle 1 is an obtuse angle and angle 2 is an acute angle. 8. trapezoids 9. 360 Practice 3-6 1. y = 31 x - 7 2. y = -2x + 12 3. y = 7x - 18 1 4. y = -2 x - 3 5. y = 16 x - 23 6. y =45 x - 2 7. y = 4x - 13 8. y = -x + 6 9. 2 4 6 x 2 4 6 x –6–4–2 –2 –4 –6 y 1 x 5 4 –6 Guided Problem Solving 3-3 1. supplementary angles 2. right angle 3. Any one of the following: Postulate 3-1, or Theorem 3-1, 3-2, 3-3 or 3-4 4. 90 5. It is congruent; Postulate 3-1 6. 90 7. a ⊥ c 8. It is true for any lineparallel to b. 9. Yes. The point is that a 11. Practice 3-4 13. 1. three 2. 180 3. right triangle 4. z = 90; Because it is given in the figure that BD ⊥ AC. 5. Theorem 3-12, the Triangle Angle-Sum Theorem 6. x = 38 7. y = 36 8. #ABD is a 36-54-90 right triangle. #BCD is a 38-52-90 right triangle. 9. 74 10. #ABCis a 52-54-74 acute triangle. 11. Yes, all three are acute angles, with &ABC visibly larger than &A and &C. 12. &BCD Practice 3-5 1. x = 120; y = 60 2. n = 5137 3. a = 108; b = 72 4. 109 5. 133 6. 129 7. 129 8. 47 9. 127 10. 30 11. 150 12. 6 13. 5 14. 8 15. BEDC 16. FAE 17. FAE and BAE 18. ABCDE 19.20 70 Geometry x 4 2 –6–4 –2 –2 –4 –6 17. y 6 y 2x 3 4 2 –6–4–2 –2 y 16. y 3x 5 2 4 6 x y 2 6 y x2 3 4 4 x 2 4 6 –6 2 4 6 x –2 –2 –4 –6 18. y –4 –6 14. 2 4 6 x 6 4 y 5x 4 2 –6–4–2 2 4 6 x –6–4–2 –4 –6 y 15. 6 4 2 2 4 6 6 y 2x 7 4 2 –6–4 –2 –2 –4 –6 y yx1 12. y 6 4 2 –6–4–2 –2 –4 –6 1. 125 2. 69 3.143 4. 129 5. 140 6. 136 7. x = 35; y = 145; z = 25 8. a = 55; b = 97; c = 83 9. v = 118; w = 37; t = 62 10. 50 11. 88 12. m3 = 22; m4 = 22; m5 = 88 13. 57.1 14. 136 15. m1 = 33; m2 = 52 16. isosceles 17. obtuse scalene 18. right scalene 19. obtuse isosceles 20. equiangular equilateral Guided ProblemSolving 3-4 y 1 x 2 3 transversal cannot be perpendicular to just one of two parallel lines. It has to be perpendicular to both, or else to neither. y 6 4 2 y 4x 1 –6–4–2 –2 l3 10. y 6 4 2 All rights reserved. l2 Guided Problem Solving 3-5 2 4 6 x y 6 4 2 2 6x –6–4–2 –2 1 –4 y 2 x 3 –6 All-In-One AnswersVersion A © Pearson Education, Inc., publishing as Pearson Prentice Hall. or boulevard), and adjacent sides must be of opposite type. 5. Not necessarily true. If there are more than three avenues and more than three boulevards, there will be some blocks bordered by neither Center City Boulevard norFounders Avenue. 6. a ⊥ e 7. a || e 8. a || e 9. a ⊥ e 10. a ⊥ e 11. a || e 12. If the number of ⊥ statements is even, then 1 || n. If it is odd, then 1 ⊥ n. 13. The proof can instead use alternate interior angles or alternate exterior angles (if the angles are congruent, the lines are parallel) or same-side interioror same-side exterior angles (if the angles are supplementary, the lines are parallel). 14. It is possible. (continued) Geometry: All-In-One Answers Version A (continued) 19. 20. y 6 4 2 x 2 –6–4 y 2x 2 4 6 x –2 –4 –6 43. y 2 4 6 x –6–4–2 –2 –4 –6 22. y All rights reserved. –6–4–2 –2 –4 –6 6 4 2 2 4 6 x y 64 2 – y 3x 2 2 –6–4–2 –2 –4 –6 25. y = –3x + 13 28. y = -21 x - 12 31. y = -15 x - 65 y = -11 34. x = 36. x = -1; y = 8 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 37. –4 –6 24. 2 4 6 x 38. 45a. m = $0.10 45b. the amount of money the worker is paid for each box loaded onto the truck45c. b = $3.90 45d. the base amount the worker is paid per hour 46. y = -25 x + 8 –6 –4 –2 2 4 6 x –2 –4 –6 (0, –1) y 6 4 1 (– , 0) 2 (0, 2) 2 2 4 6 x –6–4–2 (6, 0) –4 –6 42. 6 4 2 (4, 0) 2 4 6 x –6–4–2 –2 –4 (0, –4) –6 All-In-One Answers Version A y 6 4 2 (0, 1) (8, 0) –6–4–2 –2 –4 –6 Guided ProblemSolving 3-6 1. y B 2 O A 2 x C 2 y 2y 4 2 2 4 6 x y –2 –2 2 4 6 x –4 –6 2. m 5 x22 2 x11 3. y - y1 = m(x-x1) 4. Slope of (–2, 0) 6 6 (0, 6) 4 2 41. 2 4 6 8 –4 –6 2 y 40. y (9, 0) x 2 4 6 x –6–4–2 –2 –4 –6 –6–4–2 2 6 x –2 –4 –6 –8 –10 –12 (0, 12) –6–4–2 –2 –4 –6 x 2.5 26. y = x + 4 27. y = 21 x - 3 29. y = 2x+ 4 30. y = 13 x + 4 32. y = -6x + 45 33. x = 2; 0; y = 2 35. x = -4; y = -4 6 4 (4, 0) 2 (0, 6) y 6 4 2 y 39. 2 4 6 x –6–4–2 y 5 23. yx –6 –4–2 –2 6 4 (–3, 0) 2 y 6 4 2 y 6 4 2 12 (0, 12) 10 –6 21. 44. y 6 4 2 4 6 8 x å å AB = 5 ; slope of BC = - 5 . The absolute values of the 2 2 slopes are the same, but one slopeis positive and the other is negative. 5. Point-slope form: y - 0 = 5 (x - 0); 2 5 slope-intercept form: y = x 6. Point-slope form: 2 y - 5 = - 5 (x - 2) or y - 0 = - 5 (x - 4); 2 2 å 5 slope-intercept form: y = - x + 10 7. Of line AB : (0, 0); 2 å of line BC : (0, 10) 8. #ABC appears to be an isosceles triangle, which isconsistent with a horizontal base and two remiaining sides having slopes of equal magnitude and opposite sign. 9. Slope = 0; y = 0; y-intercept = (0, 0) just as for line å AB (they intersect on the y-axis). Practice 3-7 1. neither; 3 2 13, 3 ? 13 2 -1 2. perpendicular; 2 1 2 2 ? -2 = -1 3. parallel; -3 = -3 4.parallel; -1 = -1 5. perpendicular; y = 2 is a horizontal line, x = 0 is a vertical line 6. parallel; -21 = -12 7. neither; 1 2 18, 1 ? 81 2 -1 8. parallel; -32 = -23 9. perpendicular; -1 ? 1 = -1 10. neither; 21 2 7 7 -53, 21 ? –53 2 -1 11. neither; -23 2 -12 , -32 ? -12 2 -1 Geometry 71 Geometry: All-In-One AnswersVersion A neither; 6 2 -51, 6 ? -51 2 -1 neither; 29 2 4, 29 ? 4 2 -1 parallel; 21 = 21 15. y = 23 x y = -43 x + 24 y y = -x - 3 6 4 y = 35 x + 6 2 y=0 4 6 x –6–4 L y = 2x - 4 –4 y = 2x 10. Sample: b a 11. Sample: Guided Problem Solving 3-7 1. b y c K H 4 x G 4 All rights reserved. 12. 13. 14. 16. 17. 18. 19. 20.21. (continued) O 12. 4 4 b 2. a right angle 3. m1 m2 = -1 4. sides GH and GK b 5. Slope of GH = 35 ; slope of GK = - 38 b 6. Product = - 85 -1. Sides GH and GK are not perpendicular. 7. #GHK has no pair of perpendicular sides. It is not a right triangle. 8. No 9. &HGK; approximately 80 10. Slope of LM =72 and slope of LN = - 27 . The product © Pearson Education, Inc., publishing as Pearson Prentice Hall. of the slopes is -1, so LM and LN are perpendicular. b 13. Practice 3-8 1.–3. 4.–6. a Q T c 14. Sample: 7.–9. K b 72 Geometry c All-In-One Answers Version A Geometry: All-In-One Answers Version A(continued) 8. Line AB is perpendicular to line DF. 9. Angle ABC and angle ABD are complementary. 10. Angle 2 is a right angle, or the measure of angle 2 is 90°. 11. Sample answer: å å CB GD, m/BAF = m/GFA 15. a 3D: Visual Vocabulary Practice/High-Use Academic Words 1. property 2. conclusion 3.describe 4. formula 5. measure 6. approximate 7. compare 8. contradiction 9. pattern a Guided Problem Solving 3-8 All rights reserved. 1. a line segment of length c 2. Construct a quadrilateral with one pair of parallel sides of length c, and then examine the other pair. 3. The procedure is given on p. 181 ofthe text. 3E: Vocabulary Check Transversal: A line that intersects two coplanar lines in two points. Alternate interior angles: Nonadjacent interior angles that lie on opposite sides of the transversal. Same-side interior angles: Interior angles that lie on the m l 4. Adjust the compass to exactly span linesegment c, end to end. Then tighten down the compass adjustment as necessary. same side of a transversal between two lines. Corresponding angles: Angles that lie on the same side of a transversal between two lines, in corresponding positions. Flow proof: A convincing argument that uses deductivereasoning, in which arrows show the logical connections between the statements. 5. m 3F: Vocabulary Review l 1. C 2. E 3. D 4. B 5. A 6. F 7. K 8. H 9. L 10. G 11. I 12. J Chapter 4 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 6. m Practice 4-1 l 7. They appear to be both congruentand parallel. 8. The answers to Step 7 are confirmed. 9. yes; a parallelogram 3A: Graphic Organizer 1. Parallel and Perpendicular Lines 2. Answers may vary. Sample: properties of parallel lines; finding the measures of angles in triangles; classifying polygons; and graphing lines 3. Check students’ work.3B: Reading Comprehension 1. 14 spaces 2. 6 spaces 3. 60° 4. corresponding 5. $7000; $480 6. the width of the stalls, 10 ft 7. b 3C: Reading/Writing Math Symbols å å 1. m # n 2. m/1 + m/2 = 180 3. AB CD 4. m/MNP + m/MNQ = 90 5. /3 /EFD 6. Line 1 is parallel to line 2. 7. The measure of angle ABC isequal to the measure of angle XYZ. All-In-One Answers Version A 1. m1 = 110; m2 = 120 2. m3 = 90; m4 = 135 3. m5 = 140; m6 = 90; m7 = 40; m8 = 90 4. CA JS , AT SD , CT JD 5. C J, A S, T D 6. WZ JM , WX JK , XY KL , ZY ML 7. W J, X K, Y L, Z M 8. Yes; GHJ IHJ by Theorem 4-1 and by theReflexive Property of . Therefore, GHJ IHJ by the definition of triangles. 9. No; QSR TSV because vertical angles are congruent, and QRS TVS by Theorem 4-1, but none of the sides are necessarily congruent. 10a. Given 10b. Vertical angles are . 10c. Theorem 4-1 10d. Given 10e. Definition of trianglesGuided Problem Solving 4-1 1. right triangles 2. m&A = 45, m&B = m&L = 90, and AB = 4 in. 3. #ABC is congruent to #KLM means corresponding sides and angles are congruent. 4. x and t 5. 45 6. m&K = m&M = 45 7. 3x = 45 8. x = 15 9. 4 10. 2t = 4 11. t = 2 12. The angle measures indicate that the twotriangles are isosceles right triangles. This matches the appearance of the figure. 13. m&M = 60 Geometry 73 Geometry: All-In-One Answers Version A (continued) Practice 4-2 Practice 4-4 1. ADB CDB by SAS 2. not possible 3. not possible 4. TUS XWV by SSS 5. not possible 6. DEC GHF by SAS 7.MKL KMJ by SAS 8. PRN PRQ by SSS 9. not possible 10. C 11. AB and BC 12. A and B 13. AC 14a. Given 14b. Reflexive Property of Congruence 14c. SAS Postulate 15. Statements Reasons 1. EF FG , DF FH 1. Given 1. BD is a common side, so ADB CDB by SAS, and A C by CPCTC. 2. FH is acommon side, so FHE Guided Problem Solving 4-2 1. IS > OS and SP bisects &ISO. 2. Prove whatever additional facts can be proven about #ISP and #OSP, based on the given information. 3. IS > OS 4. &ISP &PSO 5. SP 6. #ISP #OSP by Postulate 4-2, the Side-AngleSide (SAS) Postulate 7. It does notmatter. The Side-AngleSide Postulate applies whether or not they are collinear. 8. It does follow, because #ISP #OSP and because IP and PQ are corresponding parts. Practice 4-3 1. not possible 2. ASA Postulate 3. AAS Theorem 4. AAS Theorem 5. not possible 6. not possible 7. ASA Postulate 8. notpossible 9. AAS Theorem 10. Statements Reasons 1. K M, KL ML 2. JLK PLM 3. JKL PML 11. 1. Given 2. Vertical s are . 3. ASA Postulate Q S Given TRS RTQ QRT STR Given AAS Theorem RT TR Reflexive Property of 12. BC EF 13. KHJ HKG or KJH HGK 14. NLM NPQ Guided Problem Solving 4-31. corresponding angles and alternate interior angles 2. &EAB and &DBC 3. &EBA and &DCB. 4. &EAB and &DBC 5. &EAB > &DBC, AE > BD, and &E > &D 6. #AEB > #BDC by Postulate 4-3, the AngleSide-Angle (ASA) Postulate 7. Yes; Theorem 4-2, the Angle-Angle-Side (AAS) Theorem; &EBA >&DCB 8. No, because now there is no way to demonstrate a second Guided Problem Solving 4-4 1. A compass with a fixed setting was used to draw two circular arcs, both centered at point P but crossing in different locations, which were labeled A and B. The compass was used again, with a widersetting, to draw two intersecting circular arcs, one centered at A and one at B. The point at å which the new arcs intersected was labeled C. Finally, line CP was drawn. 2. Find equal lengths or distances and explain å why CP is perpendicular to . 3. #ACP and #BCP 4. AP = PB and AC = BC. 5. #APC >#BPC, by Postulate 4-1, the Side-Side-Side (SSS) Postulate 6. &APC > &BPC by CPCTC 7. Since &APC > &BPC, m&APC = m&BPC and m&APC + m&BPC = 180, it follows that m&APC = m&BPC = 90. 8. from the definition of perpendicular and the fact that m&APC = m&BPC = 90 9. The distances donot matter, so long as AP = BP and AC = BC. That is what is required in order that #APC > #BPC. 10. Draw a line, and use the construction technique of the problem to construct a second line perpendicular to the first. Then do the same thing again to construct a third line perpendicular to the second line.The first and third lines will be parallel, by Theorem 3-10. Practice 4-5 1. x = 35; y = 35 2. x = 80; y = 90 3. t = 150 4. r = 45; s = 45 5. x = 55; y = 70; z = 125 6. a = 132; b = 36; c = 60 7. x = 6 8. a = 30; b = 30; c = 75 9. z = 120 10. AD ; D F 11. GA ; ACG AGC 12. KJ ; KIJ KJI 13. DC ; CDE CED 14. BA ;ABJ AJB 15. CB ; BCH BHC 16. 130 17. 65 18. 130 19. 90 20. x = 70; y = 55 21. x = 70; y = 20 22. x = 45; y = 45 pair of congruent sides, nor a second pair of congruent angles. 74 Geometry All-In-One Answers Version A All rights reserved. 2. Vertical s are . 3. SAS Postulate UVX WVX by SSS, and U Wby CPCTC. 6. ZAY and CAB are vertical angles, so ABC AYZ by ASA. ZA AC by CPCTC. 7. EG is a common side, so DEG FEG by SAS, and FG DG by CPCTC. 8. JKH and LKM are vertical angles, so HJK MLK by AAS, and JK KL by CPCTC. 9. PR is a common side, so PNR RQP by SSS, and N Q byCPCTC. 10. First, show that ACB and ECD are vertical angles. Then, show ABC EDC by ASA. Last, show A E by CPCTC. 11. First, show FH as a common side. Then, show JFH GHF by SAS. Last, show FG JH by CPCTC. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2. DFE HFG 3.DFE HFG HFG by ASA, and HE FG by CPCTC. 3. KLJ PMN by ASA, so K P by CPCTC. 4. QS is a common side, so QTS SRQ by AAS. QST SQR by CPCTC. 5. VX is a common side, so Geometry: All-In-One Answers Version A (continued) Guided Problem Solving 4-5 GI JI 4. 1. One angle is obtuse.The other two angles are acute and congruent. 2. Highlight an obtuse isosceles angle and find its Given angle measures, then find all the other angle measures represented in the figure. 3. Possible answer: HI HI IHG IHJ Reflexive Property of HL Theorem GHI JHI GHI and JHI are right s. Given Theorem2-5 mGHI + mJHI = 180 Angle Addition Postulate All rights reserved. 4. 60 5. 30, because the measure of each base angle is half the measure of an angle of the equilateral triangle. 6. 120° because the sum of the angles of the highlighted triangle must equal 180°. 7. The other measures are 90° and 150°.Examples: 5. RS VW 6. none 7. mC and mF = 90 8. GH JH 9. LN PR 10. ST UV or SV UT 11. mA and mX = 90 12. mF and mD = 90 13. GI # JH Guided Problem Solving 4-6 1. Two congruent right triangles. Each one has a leg and a hypotenuse labeled with a variable expression. 2. the values of x and yfor which the triangles are congruent by HL 3. The two shorter legs are congruent. 4. x = y + 1 5. The hypotenuses are congruent. 6. x + 3 = 3y 7. x = 3; y = 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 8. hypotenuse shorter leg = 2; yes, this matches the figure. 9. The 8. Yes; 3 120 =360. 9. Answers may vary. solution remains the same: x = 3 and y = 2. The reason is that one is still solving the same two equations, x = y + 1 and x + 3 = 3y. Practice 4-6 Practice 4-7 1. Statements 1. AB # BC , ED # FE 2. B, E are right s. 3. AC FD , AB ED 4. ABC DEF 2. Statements 1. P, R are right s.2. PS QR 3. SQ QS 4. PQS RSQ 3. Reasons 1. Given 2. Perpendicular lines form right s. 3. Given 4. HL Theorem Reasons 1. Given 2. Given 3. Reflexive Property of 4. HL Theorem MJN and MJK are right s. 1. ZWX YXW; SAS 2. ABC DCB; ASA 3. EJG FKH; ASA 4. LNP LMO; SAS 5. ADF BGE; SAS 6.UVY VUX; ASA 7. B A C B C common side: BC F common side: FG D 8. F G E G H 9. common angle: L L Perpendicular lines form right s. L N K MN MK MJN MJK Given HL Theorem M J MJ MJ Reflexive Property of All-In-One Answers Version A Geometry 75 Geometry: All-In-One Answers Version AReasons 1. Given 2. Given 3. Perpendicular lines form right s. 4. Reflexive Property of 5. BYA CXA 5. ASA Postulate 11. Sample: Because FH GE, HFG EGF, and FG GF, then FGE GFH by SAS. Thus, FE GH by CPCTC and EH HE, then GEH FHE by SSS. Guided Problem Solving 4-7 4E: VocabularyCheck Angle: Formed by two rays with the same endpoint. Congruent angles: Angles that have the same measure. Congruent segments: Segments that have the same length. Corresponding polygons: Polygons that have corresponding sides congruent and corresponding angles congruent. CPCTC: Anabbreviation for “corresponding parts of congruent triangles are congruent.” 4F: Vocabulary Review Puzzle 1. postulate 2. hypotenuse 3. angle 4. vertex 5. side 6. leg 7. perpendicular 8. polygon 9. supplementary 10. parallel 11. corresponding 1. The figure, a list of parallel and perpendicular pairs of sides,and one known angle measure, namely m&A = 56 2. nine 3. They are congruent and have equal measures. 4. m&A = m&1 = m&2 = 56 5. m&4 = 90 6. m&3 = 34 7. m&DCE = 56 8. m&5 = 22 9. m&FCG = 90 10. m&6 = 34 11. m&7 = 34, m&8 = 68, and m&9 = 112 12. m&9 = 56 + 56 = 112 13. m&FIC =180 -(m&2 + m&3) = 90; m&DHC = m&4 = 90; m&FJC = 180 - m&9 = 68; m&BIG = m&FIC = 90 Chapter 5 4A: Graphic Organizer Guided Problem Solving 5-1 1. Congruent Triangles 2. Answers may vary. Sample: congruent figures; triangle congruence by SSS, SAS, ASA, and AAS; proving parts oftriangles congruent; the Isosceles Triangle Theorem 3. Check students’ work. 4B: Reading Comprehension 1. Yes. Using the Isosceles Triangle Theorem, /W /Y. It is given that WX YX and WU YV. Therefore nWUX nYVX by SAS. 2. There is not enough information. You need to know if AC EC, if /A /E, orif /B /D. 3. a 4C: Reading/Writing Math Symbols Practice 5-1 1a. 8 cm 1b. 16 cm 1c. 14 cm 2a. 22.5 in. 2b. 15.5 in. 2c. 15.5 in. 3a. 9.5 cm 3b. 17.5 cm 3c. 14.5 cm 4. 17 5. 20.5 6. 7 7. 128 19 8. 42 9. 16.5 10a. 18 10b. 61 11. GH 6 AC, HI 6 BA, GI 6 BC 12. PR 6 YZ, PQ 6 XZ, XY 6 RQ 1. 30 units 2. Thethree sides of the large triangle are each bisected by intersections with the two line segments lying in the interior of the large triangle. 3. the value of x 4. They are called midsegments. 5. They are parallel, and the side labeled 30 is half the length of the side labeled x. 6. x = 60 7. Yes; the side labeled xappears to be about twice as long as the side labeled 30. 8. No. Those lengths are not fixed by the given information. (The triangle could be vertically stretched or shrunk without changing the lengths of the labeled sides.) All one can say is that the midsegment is half as long as the side it is parallel to.Practice 5-2 1. Angle-Angle-Side 2. triangle XYZ 3. angle PQR 4. line segment BD 5. line ST 6. ray WX 7. hypotenuseleg 8. line 3 9. angle 6 10. Angle-Side-Angle 1. WY is the perpendicular bisector of XZ. 2. 4 3. 7.5 4. 9 5. right triangle 6. 5 7. 17 8. 17 9. equidistant 10. isosceles triangle ) 11. 3.5 12. 2113. 21 14. right triangle 15. JP is the bisector of LJN. 16. 9) 17. 45 18. 45 19. 14 20. Sample: Point M lies on JP . 21. right 4D: Visual Vocabulary Practice isosceles triangle 1. theorem 2. congruent polygons 3. base angle of an isosceles triangle 4. CPCTC 5. postulate 6. vertex angle of an isosceles triangle7. corollary 8. base of an isosceles triangle 9. legs of an isosceles triangle 76 Geometry All rights reserved. Statements 1. AX AY 2. CX # AB, BY # AC 3. mCXA and mBYA = 90 4. A A All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 10. Sample: (continued)Geometry: All-In-One Answers Version A (continued) Guided Problem Solving 5-2 1. < y m A C 4 B x 4 O All rights reserved. 2. See answer to Step 1, above. 3. See answer to Step 1, above. 4. Plot a point and explain why it lies on the bisector of the angle at the origin. 5. line : y 5 23 x 1 25 ; line m: 4 2 x= 10 6. C(10, 5) 7. CA = CB = 5; yes 8. Theorem 5-5, the Converse of the Angle Bisector Theorem 9. m&AOC = m&BOC < 27 10. Draw , m, and C, then draw OC. Since OA = OB = 10, it follows that #OAC > #OBC, by HL. Then CA > CB and &AOC > &BOC by CPCTC. Practice 5-3 1. (-2, 2) 2. (4, 0) 3. (2,1) 4. Check students’ work. The final result of the construction is shown. B © Pearson Education, Inc., publishing as Pearson Prentice Hall. A C 5. altitude 6. median 7. none of these 8. perpendicular bisector 9. angle bisector 10. altitude 11a. (2, 0) 11b. (-2, -2) 12a. (0, 0) 12b. (3, -4) 13a. (0, 0) 13b. (0, 3)Guided Problem Solving 5-3 1. the figure and a proof with some parts left blank 2. Fill in the blanks. 3. AB 4. Theorem 5-2, the Perpendicular Bisector Theorem 5. BC; XC 6. the Transitive Property of Equality 7. Perpendicular Bisector. (This converse is Theorem 5-3.) 8. The point of the proof is todemonstrate that n runs through point X. It would not be appropriate to show that fact as already given in the figure. 9. Nothing essential would change. Point X would lie outside #ABC å (below BC ), but the proof woud run just the same. Practice 5-4 1. I and III 2. I and II 3. The angle measure is not 65. 4.Tina does not have her driver’s license. 5. The figure does not have eight sides. 6. The restaurant is open on Sunday. 7. ABC is congruent to XYZ. 8. mY 50 9a. If Toronto; true. 11. Assume that mA 2 mB. 12. Assume that TUVW is not a trapezoid. 13. Assume that LM does not intersect NO. 14. Assumethat FGH is not equilateral. 15. Assume that it is not sunny outside. 16. Assume that D is obtuse. 17. Assume that mA 90. This means that mA + mC 180. This, in turn, means that the sum of the angles of ABC exceeds 180, which contradicts the Triangle Angle-Sum Theorem. So the assumption that mA90 must be incorrect. Therefore, mA 90. Guided Problem Solving 5-4 1. Ice is forming on the sidewalk in front of Toni’s house. 2. Use indirect reasoning to show that the temperature of the sidewalk surface must be 32°F or lower. 3. The temperature of the sidewalk in front of Toni’s house is greater than32°F. 4. Water is liquid (ice does not form) above 32°F. 5. There is no ice forming on the sidewalk in front of Toni’s house. 6. The result from Step 5 contradicts the information identified as given in Step 1. 7. The temperature of the sidewalk in front of Toni’s house is less than or equal to 32°F. 8. If thetemperature is above 32°F, water remains liquid. This is reliably true. Converse: If water remains liquid, the temperature is above 32°F. This is not reliably true. Adding salt will cause water to remain liquid even below 32°F. 9. Suppose two people are each the world’s tallest person. Call them person A andperson B. Then person A would be taller than everyone else, including B, but by the same token B would be taller than A. It is a contradiction for two people each to be taller than the other. So it is impossible for two people each to be the World’s Tallest Person. Practice 5-5 1. M, N 2. C, D 3. S, Q 4. R, P5. A, T 6. S, A 7. yes; 4 + 7 8, 7 + 8 4, 8 + 4 7 8. no; 6 + 10 17 9. yes; 4 + 4 4 10. yes; 1 + 9 9, 9 + 9 1, 9 + 1 9 11. yes; 11 + 12 13, 12 + 13 11, 13 + 11 12 12. no; 18 + 20 40 13. no; 1.2 + 2.6 4.9 14. no; 8 21 + 9 14 18 15. no; 2. 5 + 3.5 6 16. BC, AB, AC 17. BO, BL, LO 18. RS, ST, RT 19. D, S, A 20. N, S,J 21. R, O, P 22. 3 x 11 23. 8 x 26 24. 0 x 10 25. 9 x 31 26. 2 x 14 27. 13 x 61 Guided Problem Solving 5-5 1. two triangles are not congruent, then their corresponding angles are not congruent; false. 9b. If corresponding angles are not congruent, then the triangles are not congruent; true. 10a. If you donot live in Toronto, then you do not live in Canada; false. 10b. If you do not live in Canada, then you do not live in All-In-One Answers Version A Geometry 77 Geometry: All-In-One Answers Version A 2. The side opposite the larger included angle is greater than the side opposite the smaller included angle.3. The angle opposite the larger side is greater than the angle opposite the smaller side 4. The opposite sides each have a length of nearly the sum of the other two side lengths. 5. The opposite sides are the same length. They are corresponding parts of triangles that are congruent by SAS. (continued)isosceles trapezoid, quadrilateral 8. square, rectangle, parallelogram, rhombus, quadrilateral 9. x = 7; AB = BD = DC = CA = 11 10. m = 9; s = 42; ON = LM = 26; OL = MN = 43 11. f = 5; g = 11; FG = GH = HI = IF = 17 12. parallelogram 13. rectangle 14. kite 15. parallelogram Guided Problem Solving 6-1Sample: midsegments of triangles; bisectors in triangles; concurrent lines, medians, and altitudes; and inverses, contrapositives, and indirect reasoning 3. Check students’ work. 5B: Reading Comprehension 1. The width of the tar pit is 10 meters. 2. b 1. a labeled figure, which shows an isosceles

trapezoid 2. The nonparallel sides are congruent. 3. the measures of the angles and the lengths of the sides 4. m&G = c 5. c + (4c - 20) = 180 6. 40 7. m&D = m&G = 40, m&E = m&F = 140 8. a - 4 = 11 9. 15 10. DE = FG = 11, EF = 15, DG = 32 11. 40 + 40 + 140 + 140 = 360 = (4 - 2)180 12. m&D = m&G= 39, m&E = m&F = 141 Practice 6-2 1. L 2. F 3. O 4. G 5. A 6. I 7. M 8. H 9. E 10. K 11. B 12. D 13. N 14. J 15. C 1. 15 2. 32 3. 7 4. 8 5. 12 6. 9 7. 8 8. 3 58 9. 54 10. 34 11. 54 12. 34 13. 100 14. 40; 140; 40 15. 70; 110; 70 16. 113; 45; 22 17. 115; 15; 50 18. 55; 105; 55 19. 61; 72; 108; 32 20. 32; 98; 5021. 16 22. 35 23. 28 24. 4 5D: Visual Vocabulary Practice Guided Problem Solving 6-2 1. median 2. negation 3. circumcenter 4. contrapositive 5. centroid 6. equivalent statements 7. incenter 8. inverse 9. altitude 1. the ratio of two different angle measures in a parallelogram 2. The consecutive angles aresupplementary. 3. 9x 5C: Reading/Writing Math Symbols 5E: Vocabulary Check Midpoint: A point that divides a line segment into two congruent segments. Midsegment of a triangle: The segment that joins the midpoints of two sides of a triangle. Proof: A convincing argument that uses deductivereasoning. Coordinate proof: A proof in which a figure is drawn on a coordinate plane and the formulas for slope, midpoint, and distance are used to prove properties of the figure. Distance from a point to a line: The length of the perpendicular segment from the point to the line. 5F: Vocabulary Review 1.altitude 2. line 3. median 4. negation 5. contrapositive 6. incenter 7. orthocenter 8. slope-intercept 9. exterior 10. obtuse 11. alternate interior 12. centroid 13. equivalent 14. right 15. parallel Chapter 6 Practice 6-1 1. parallelogram 2. rectangle 3. quadrilateral 4. parallelogram, quadrilateral 5. kite,quadrilateral 6. rectangle, parallelogram, quadrilateral 7. trapezoid, 78 Geometry All rights reserved. 1. Relationships Within Triangles 2. Answers may vary. x 4. the measures of the angles 5. The angles are supplementary angles, because they are consecutive. 6. x + 9x = 180 7. 18 and 162 8. No; thelengths of the sides are irrelevant in this problem. 9. 30 and 150 Practice 6-3 1. no 2. yes 3. yes 4. no 5. yes 6. yes 7. x = 2; y = 3 8. x = 6; y = 3 9. x = 64; y = 10 10. x = 8; the figure is a because both pairs of opposite sides are congruent. 11. x = 40; the figure is not a because one pair of opposite angles isnot congruent. 12. x = 25; the figure is a because the congruent opposite sides are 6 by the Converse of the Alternate Interior Angles Theorem. 13. Yes; the diagonals bisect each other. 14. No; the congruent opposite sides do not have to be 6. 15. No; the figure could be a trapezoid. 16. Yes; both pairs ofopposite sides are congruent. 17. Yes; both pairs of opposite sides are 6 by the converse of the Alternate Interior Angles Theorem. 18. No; only one pair of opposite angles is congruent. 19. Yes; one pair of opposite sides is both congruent and 6. 20. No; only one pair of opposite sides is congruent. All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 5A: Graphic Organizer Geometry: All-In-One Answers Version A (continued) Guided Problem Solving 6-3 Guided Problem Solving 6-5 1. a labeled figure, which shows a quadrilateral that appears to be aparallelogram 2. The consecutive angles are supplementary. 3. find values for x and y which make the quadrilateral a parallelogram 4. m&A + m&D = 180, so 1. Isosceles trapezoid ABCD with AB > DC 2. &B > &C and &BAD > &D 3. AB > DC is Given. DC > AE because opposite sides of a parallelogramare congruent (Theorem 6-1). AB > AE is from the Transitive Property of Congruency. 4. Isosceles; >; because base angles of an isosceles triangle are congruent. 5. &1 > &C because corresponding angles on a transversal of two parallel lines are congruent. 6. &B > &C by the Transitive Property ofCongruency. 7. &BAD is a same-side interior angle with &B, and &D is a same-side interior angle with &C. 8. This is not a problem, because for AD > BC there is a similar proof with a line segment drawn from B to a point E lying on AD. 9. The two drawn segments can be shown to be congruent, and thenone has two congruent right triangles by the HL Theorem. &B > &C follows by CPCTC and &BAD > &D because they are supplements of congruent angles. that &A and &D meet the requirements for same-side interior angles on the transversal of two parallel lines (Theorems 3-2 and 3-6). 5. &B > &D, byTheorem 6-2. 6. 3x + 10 + 5y = 180; 8x + 5 = 5y 7. x = 15, y = 25 8. m&A = m&C = 55 and m&B = m&D = 125, which matches the appearance of the figure. 9. (3x + 10) + (8x + 5) = 180; yes © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Practice 6-4 1a. rhombus 1b.72; 54; 54; 72 2a. rectangle 2b. 72; 36; 18; 144 3a. rectangle 3b. 37; 53; 106; 74 4a. rhombus 4b. 59; 90; 90; 59 5a. rectangle 5b. 60; 30; 60; 30 6a. rhombus 6b. 22; 68; 68; 90 7. Yes; the parallelogram is a rhombus. 8. Possible; opposite angles are congruent in a parallelogram. 9. Impossible; if thediagonals are perpendicular, then the parallelogram should be a rhombus, but the sides are not of equal length. 10. x = 7; HJ = 7; IK = 7 11. x = 7; HJ = 26; IK = 26 12. x = 6; HJ = 25; IK = 25 13. x = -3; HJ = 13; IK = 13 14a. 90; 90; 29; 29 14b. 288 cm2 15a. 70; 90; 70; 70 15b. 88 in.2 16a. 38; 90; 90; 3816b. 260 m2 17. possible 18. Impossible; because opposite angles are congruent and supplementary, for the figure to be a parallelogram they must measure 90, the figure therefore must be a rectangle. Guided Problem Solving 6-4 1. A labeled figure, which shows a parallelogram. One angle is a rightangle, and two adjacent sides are congruent. Algebraic expressions are given for the lengths of three line segments. 2. diagonals 3. Find the values of x and y. 4. It is a square. Theorem 6-1 and the fact that AB > AD imply that all four sides are congruent. Theorems 3-11 and 6-2 plus the fact that m&B =90 imply that all four angles are right angles. 5. congruent; bisect 6. 4x - y + 1 = (2x - 1) + (3y + 5); 2x - 1 = 3y + 5 7. x = 7 1 ; y = 3 8. It was not necessary to 2 know AB > AD, but it was necessary to know m&B = 90. The key fact, which enables the use of Theorem 6-11 in addition to Theorem 6-3, is thatABCD is a rectangle. It does not matter whether all four sides are congruent. 9. 40 Practice 6-5 1. 118; 62 2. 99; 81 3. 59; 121 4. 96; 84 5. 101; 79 6. 67; 113 7. x = 4 8. x = 16; y = 116 9. x = 1 10. 105.5; 105.5 11. 90; 25 12. 118; 118 13. 90; 63; 63 14. 107; 107 15. 90; 51; 39 16. x = 8 17. x = 7 18. x = 28; y= 32 All-In-One Answers Version A Practice 6-6 1. (1.5a, 2b); a 2. (1.5a, b); " a2 1 4b2 3. (0.5a, 0); a 4. (0.5a, b); " a2 1 4b2 5. 0 6. 1 7. -12 8. 2 9. 2b 3a 2b 2b 10. -2b 11. 12. 13. E( a, 3 b ); I(4 a, 0) 3a 3a 3a 14. O(3a, 2b); M(3a, -2b); E(-3a, -2b) 15. D(4a, b); I(3a, 0) 16. T(0, 2b); A(a, 4b); L(2a, 2b) 17. (-4a, b) 18. (-b, 0) Guided Problem Solving 6-6 1. a rhombus with coordinates given for two vertices 2. A rhombus is a parallelogram with four congruent sides. 3. the coordinates of the other two vertices 4. They are the diagonals. 5. They bisect each other. 6. W(-2r, 0), Z(0, -2t) 7. No; neither Theorem 6-3nor any other theorem or result would apply. 8. Slope of WX = 0 and slope of YZ is undefined. This confirms Theorem 6-10, which says that the diagonals of a rhombus are perpendicular. Practice 6-7 p p2 p q 1b. y = mx + b; q = q (p) + b; b = q - q ; p p p2 y = q x + q - q 1c. x = r + p 1d. y = q (r + p) + 2 2 2rp rp p p p q - q ; y = q + q + q - q ; y = q + q; intersection rp at (r + p, q + q) 1e. qr 1f. (r, q) 1g. y = mx + b; 2 2 q = qr (r) + b; b = q - rq ; y = rqx + q - rq 2 2 2 rp 1h. y = qr (r + p) + q - rq ; y = rq + q + q - rq ; rp rp y = q + q; intersection at (r + p, q + q) rp 1i. (r + p, q + q) 2a. (-2a, 0) 2b. (-a, b) b 3 b 2c. (-3a 2 ,2 ) 2d. a 3a. (-4a, 0) 3b. (-2a, 3a) 3c. 2 3d. (2a, -a) 3e. 12 4. The coordinates for D are (0, 2b). 1a. The coordinates for C are (2a, 0). Given these coordinates, the lengths of DC and HP can be determined: Geometry 79 Geometry: All-In-One Answers Version A DC = "(2a 2 0) 2 1 (0 2 2b) 2 = " 4a2 1 4b2;HP = "(0 2 2a) 2 1 (0 2 2b) 2 = " 4a2 1 4b2; DC = HP, so DC HP. (continued) Guided Problem Solving 6-7 the given information, only one side and one angle of the two triangles can be proven to be congruent. Another side or angle is needed. If it were given that QUSV is a parallelogram, then the proofcould be made. 3. All four sides are congruent. 4. Yes. Since EG > EG by the Reflexive Property, nEFG > nEHG by SSS. 5. b 1. kite DEFG with DE = EF with the midpoint of each side identified 2. A kite is a quadrilateral with two pairs of 6C: Reading/Writing Math Symbols adjacent sides congruent and noopposite sides congruent. 3. The midpoints are the vertices of a rectangle. 4. D(-2b, 2c), G(0, 0) 5. L(b, a + c), M(b, c), N(-b, c), K(-b, a + c) 6. Slope of KL = slope of NM = 0, slopes of KN and LM are undefined 7. Opposite sides are parallel; it is a rectangle. 8. Adjacent sides are perpendicular. 9. rightangles 10. Answers will vary. Example: a = 3, b = 2, c = 2 yields the points D(-4, 4), E(0, 6), F(4, 4), G(0, 0) with midpoints at (-2, 2), (-2, 5), (2, 5), and (2, 2). Connecting these midpoints forms a rectangle. 11. Construct DF and EG. Slope of DF = 0, so DF is horizontal. Slope of EG is undefined, so EG isvertical. 1. AH, CH, or BH 2. DG, FG, or EG 3. G 4. DE 5. rhombus 6. rectangle 7. square 8. isosceles 6D: Visual Vocabulary Practice/High-Use Adademic Words 1. solve 2. deduce 3. equivalent 4. indirect 5. equal 6. analysis 7. identify 8. convert 9. common 6E: Visual Vocabulary Check Consecutiveangles: Angles of a polygon that share a common side. Kite: A quadrilateral with two pairs of congruent adjacent sides and no opposite sides congruent. Parallelogram: A quadrilateral with two pairs of parallel sides. Rhombus: A parallelogram with four congruent sides. Trapezoid: A quadrilateral withexactly one pair of parallel sides. 6A: Graphic Organizer 1. Quadrilaterals 2. Answers may vary. Sample: classifying quadrilaterals; properties of parallelograms; proving that a quadrilateral is a parallelogram; and special parallelograms 3. Check students’ work. 6B: Reading Comprehension 1. QT > SR,QR > ST, QT 6 RS, QR 6 TV 2. No, it cannot be proven that nQTV > nSRU because with 1 2 P A R A L 4 T L 5 R E L A P O E G Z R A O I H 7 O M 8 B U S 9 C 3 I M N I C D E S N E N T 6 R O E R A M R C E U E N M A T C G Geometry I G Q U A R N 80 C B S © Pearson Education, Inc., publishing asPearson Prentice Hall. 6F: Vocabulary Review Puzzle D E 10 H Y P O T E N L T E E S R U All rights reserved. trapezoid S E All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued) Guided Problem Solving 7-3 Chapter 7 Practice 7-1 © Pearson Education, Inc., publishing asPearson Prentice Hall. All rights reserved. 1. 1 : 278 2. 18 ft by 10 ft 3. 18 ft by 14 ft 4. 18 ft by 16 ft 5. 8 ft by 8 ft 6. 3 ft by 10 ft 7. true 8. false 9. true 10. true 11. false 12. true 13. true 14. false 15. false 88 16. 12 17. 12 18. 33 19. 21 5 20. 48 21. 3 22. 6 23. 25 24. -52 25. 14 : 5 26. 3 : 2 27. 12 : 7 28. 83 729. 13 1. no; N/A 2. yes; WT, RS 3. It is a trapezoid. 4. They are congruent. 5. They are parallel. 6. They are congruent. 7. #RSZ and #TWZ 8. AA~ or Angle-Angle Similarity Postulate 9. No; there is only one pair of congruent angles. 10. yes; parallelogram, rhombus, rectangle, and square Practice 7-4Guided Problem Solving 7-1 1. 16 2. 8 3. "77 4. 2 "11 5. 10"2 6. 6"5 7. h 8. y 9. x 10. a 11. b 12. c 13. x = 6; y = 6"3 14. x = 8 "3; y = 4 "3 15. 92 16. x = 4 "5 ; y = "55 7. 0.348 8. yes 9. 21.912 Guided Problem Solving 7-4 Practice 7-2 1. #ABC, #ACD, #BCD 2. 1 3. 1; 1 4. 1 5. 1 6. 2; 1 7. 2 8. "2 9. "2 10. yes11. no (This would 42 1 or 1. ratios 2. 42,000,000 3. the denominator 1,000,000 x 5 1 4. 29,000 5. Cross-Product Property 6. 0.029 1,000,000 17. x = "3; y = "6; z = "2 18. x = 8; y = 2 "2; z = 6 "2 19. 2"15 in. require the Pythagorean Theorem.) 1. ABC XYZ, with similarity ratio 2 : 1 2. QMN RST, withsimilarity ratio 5 : 3 3. Not similar; corresponding sides are not proportional. 4. Not similar; corresponding angles are not congruent. 5. ABC KMN, with similarity ratio 4 : 7 6. Not similar; corresponding sides are not proportional. 7. I 8. O 9. J 10. NO 11. LO 12. LO 13. 3.96 ft 14. 4.8 in. 15. 3.75 cm 16. 10 m17. 32 18. 53 19. 7 21 20. 4 12 21. 53 22. 37 23. 5 Practice 7-5 16 1. BE 2. EH 3. BC 4. JD 5. JG JI 6. BE 7. 3 15 25 8. 4 9. 4 10. 25 2 11. x = 9 ; y = 4 12. 4 13. x = 6; 198 189 y = 6 14. x = 5 ; y = 5 15. 2 16. 4 17. 10 Guided Problem Solving 7-5 Guided Problem Solving 7-2 1. parallel 2. CE; BD 3. 6; 15 4.90 5. yes 6. yes 7. The sides would not be parallel. 8. They are similar triangles. 1. equal 2. 6.14 3. 2.61; 6.14 4. 19.3662; 19.2182 5. no 2.61 6. no 7. 2.3706; 2.3525 8. Since the quotients are not equal, 7A: Graphic Organizer 1. Similarity 2. Answers may vary. Sample: ratios and proportions; similarpolygons; proving triangles similar; and similarity in right triangles 3. Check students’ work. the ratios are not equal, and the bills are not similar rectangles. 9. 4.045 Practice 7-3 1. AXB RXQ because vertical angles are . A R (Given). Therefore AXB RXQ by the AA 3 XM PX Postulate. 2. Because MP LW =WA = AL = 4 , MPX LWA by the SSS Theorem. 3. QMP AMB QM 2 because vertical s are . Then, because AM = PM BM = 1 , QMP AMB by the SAS Theorem. 4. M A (Given). Because there are 180° in a triangle, mJ = 130, and J C. So MJN ACB by the AA Postulate. BX 5. Because AX = BX and CX =RX, AX CX = RX . AXB CXR because vertical angles are . Therefore AXB CXR by the SAS Theorem. 6. Because AB = BC = AB BC CA CA and XY = YZ = ZX, XY = YZ = ZX . Then ABC XYZ by the SSS Theorem. 7. 11. 20 3 12. 36 13. 33 ft 15 2 8. 55 4 9. 48 7 10. 85 3 7B: Reading Comprehension 1.DHB ACB 2. AA similarity postulate The triangles have two similar angles. 3. 1 : 2 4. 1 : 2 5 HB 6. 250 ft 7. a 5. DB AB CB 7C: Reading/Writing Math Symbols 1. no 2. no 3. yes 4. no 5. yes 6. no 7. yes, AAS 8. yes, Hypotenuse-Leg Theorem 9. yes, SAS or ASA or AAS 10. not possible 7D: VisualVocabulary Practice 1. Angle-Angle Similarity Postulate 2. golden ratio 3. Side-Side-Side Similarity Theorem 4. geometric mean 5. scale 6. Cross-Product Property 7. golden rectangle 8. simplest radical form 9. Side-Angle-Side Similarity Theorem All-In-One Answers Version A Geometry 81 Geometry: All-In-One Answers Version A (continued) 7E: Vocabulary Check Practice 8-3 Similarity ratio: The ratio of lengths of corresponding sides 1. tan E = 34 ; tan F = 34 2. 3. tan E = 52; tan F = 52 4. 8. 2.3 9. 6.4 10. 78.7 11. 15. 27 16. 68 17. 39 18. proportion is equal to the product of the means. Ratio: Acomparison of two quantities by division. Golden rectangle: A rectangle that can be divided into a square and a rectangle that is similar to the original rectangle. Scale: The ratio of any length in a scale drawing to the corresponding actual length. Guided Problem Solving 8-3 1. A B 7F: Vocabulary Review1. L 2. E 3. I 4. A 5. J 6. K 7. C 8. D 9. G 10. O 11. N 12. M 13. H 14. B 15. F Chapter 8 1. " 51 2. 4 " 7 3. 2 " 65 4. " 7 5. 2 " 21 6. 18 " 2 7. 46 in. 8. 78 ft 9. 279 cm 10. 19 m 11. acute 12. right 13. obtuse 14. right 15. obtuse 16. acute Guided Problem Solving 8-1 1. the sum of the lengths of the sides 2.Pythagorean Theorem 3. 7 cm 4. 4 cm 3 cm 5. c2 = 42 + 32 6. 5 7. 12 cm 8. perimeter of rectangle = 14 cm; yes 9. Answers will vary; example: Draw a 4 cm 3 cm grid, copy the given figure, measure the lengths with a ruler, add them together. 10. 20 cm Practice 8-2 1. x = 2; y = " 3 2. a = 4.5; b = 4.5" 3 3.c = 50"3 3 ; 3 3 d = 25" 4. 8 " 2 5. 14 " 2 6. 28" 7. 2 3 3 8. x = 15; y = 15 " 3 9. 3 " 2 10. 42 cm 11. 5.9 in. 12. 10.4 ft, 12 ft 13. 170 in.2 14. w = y = 5" 2 ; z 80 cm 2. 180 3. m&A = 2m&X 4. 90 5. base: 40 cm, height: 10 cm 6. 4 7. 4 8. 76 9. 152 10. 28 11. yes 12. 46 Practice 8-4 Practice 8-1 5"3 = 3 20 cm10"3 3 ;x = 5; 15. a = 4; b = 3 16. p = 4 " 3; q = 4 " 3; r = 8; s = 4 " 6 1. sin P = 2"710 ; cos P = 3 7 3 5 = 56 8 = 17 2. sin P = 45; cos P = "11 5 12 13 ; cos P = 13 4. sin P = 6 ; cos P "2 "2 15 = 2 ; cos P = 2 6. sin P = 17 ; cos P 3. sin P = 5. sin P 7. 64 8. 11.0 9. 7.0 10. 42 11. 7.8 12. 53 13. 6.6 14. 37 15.11.0 16. 56 17. 11.5 18. 9.8 Guided Problem Solving 8-4 1. The sides are parallel. 2. sine 3. sin 30° 5 w6 4. 3.0 5. yes 6. cosine 7. cos x° 5 34 8. 41 9. Answers may vary. Sample: cos 60° 36; sin 49° 34 10. 5.2, 2.6 Practice 8-5 1a. angle of depression from the birds to the ship 1b. angle of elevation fromthe ship to the birds 1c. angle of depression from the ship to the submarine 1d. angle of elevation from the submarine to the ship 2a. angle of depression from the plane to the person 2b. angle of elevation from the person to the plane 2c. angle of depression from the person to the sailboat 2d. angle ofelevation from the sailboat to the person 3. 116.6 ft 4. 84.8 ft 5. 46.7 ft 6. 31.2 yd 7. 127.8 m 8. 323.6 m 9a. Guided Problem Solving 8-2 1. 30°-60°-90° triangle 2. l 3. h 4. Í 3 5. 24 or 8Í 3 !3 48 6. 2 7. or 16Í 3 8. 28 ft 9. 0.28 min 10. yes !3 11. 34 ft 35 30 ft 9b. 26 ft 82 Geometry All rights reserved. Cross-Product Property: The product of the extremes of a tan E = 1; tan F = 1 12.4 5. 31.0 6. 14.1 7. 7.1 26.6 12. 71.6 13. 39 14. 72 54 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. of similar polygons. Geometry: All-In-One Answers Version A (continued) GuidedProblem Solving 8-5 14. Sample: N 1. &e = 1, &d = &4 2. congruent 3. m&e = m&d 4. 7x - 5 = 4(x + 7) 5. 11 6. 72 7. 72 8. yes 9. 44, 44 30 W Practice 8-6 1. 46.0, 46.0 2. -37.1, -19.7 3. 89.2, -80.3 4. 38.6 mi/h; 31.2º north of east 5. 134.5 m; 42.0º south of west 6. 1.3 m/sec; 38.7º south of east 7. 55º northof east 8. 20º west of south 9. 33º west of north 10a. 1, 5 All rights reserved. y 6 4 642 2 4 6 x ; 1. Check students’ work. 2. 100 100 3. 100 cos 30°; 100 sin 30° 4. 86.6; 50 5. 86.6, 50 6. 86.6, -50 7. 173.2, 0 8. 173; due east 9. yes 10. 100; due east 2 4 6x 8A: Graphic Organizer 11a. 1, -1 11b. 1. RightTriangles and Trigonometry 2. Answers may vary. Sample: the Pythagorean Theorem; special right triangles; the tangent ratio; sine and cosine ratios; angles of elevation and depression; vectors 3. Check students’ work. y 6 4 4 6x 642 2 4 6 8B: Reading Comprehension 1. A 2. J 3. B 4. J 5. B 6. B 7. b 8C:Reading/Writing Math Symbols 1. F 2. G 3. D 4. A 5. C 6. B 7. H 8. E 5 10. #ABC , #XYZ 11. m&A < 52° 9. sin21 A 5 12 y 12a. -7, -1 12b. © Pearson Education, Inc., publishing as Pearson Prentice Hall. S Guided Problem Solving 8-6 y 10b. E 6 4 2 42 4 6 7 12. tan Z 5 24 2 4 6x 8D: Visual VocabularyPractice 1. 30°-60°-90° triangle 2. inverse of tangent 3. congruent sides 4. tangent 5. Pythagorean Theorem 6. hypotenuse 7. 45°-45°-90° triangle 8. Pythagorean triple 9. obtuse triangle 13. Sample: 8E: Vocabulary Check N Obtuse triangle: A triangle with one angle whose measure is between 90 and 180.W 48 E Isosceles triangle: A triangle that has at least two congruent sides. Hypotenuse: The side opposite the right angle in a right S triangle. Right triangle: A triangle that contains one right angle. Pythagorean triple: A set of three nonzero whole numbers a, b, and c that satisfy the equation a2 + b2 = c2.All-In-One Answers Version A Geometry 83 Geometry: All-In-One Answers Version A (continued) J A T N A T L U S E R Y G D H P G H S N L P U R F S I M E E M Q O I I Y H G E A K U R R E L O F L S B T N T N L O N Q G M A M M D M T P B K V I X E X X E C S Q E A L V Y D W M E T D O T S B UN T P X L T Z I T O R N I F A D R Y R L P A H S R P S Z S E O E E E J I I M T A H Y M C O N E D C L M Z C R D E G H E D P I L U T I N E B M J C V O R F P S U T A B A F N V E C T O R Y O O R I N I V G N T M A D D K E B C O N G R U E N T G H N T B R A A N G L E O F D E P R E S S I O N RA E U A R A X L L Y T N E J O B M C O O V Z L P R O P O R T I O N Practice 9-1 1. No; the triangles are not the same size. 2. Yes; the hexagons are the same shape and size. 3. Yes; the ovals are the same shape and size. 4a. C and F 4b. CD and CD, DE and DE, EF and EF, CF and CF 5a. M and N5b. MN and MN, NO and NO, MO and MO 6. (x, y) S (x - 2, y - 4) 7. (x, y) S (x - 2, y - 2) 8. (x, y) S (x - 3, y - 1) 9. (x, y) S (x + 4, y - 2) 10. (x, y) S (x - 5, y + 1) 11. (x, y) S (x + 2, y + 2) 12. W(-2, 2), X(-1, 4), Y(3, 3), Z(2, 1) 13. J(-5, 0), K(-3, 4), L(-3, -2) 14. M(3, -2), N(6, -2), P(7, -7), Q(4, -6) 15. (x, y) S (x +4.2, y + 11.2) 16. (x, y) S (x + 13, y - 13) 17. (x, y) S (x,y) 18. (x, y) S (x + 3, y + 3) 19a. P(-3, -1) 19b. P(0, 8), N(-5, 2), Q(2, 3) Guided Problem Solving 9-1 1. the four vertices of a preimage and one of the vertices of the image 2. Graph the image and preimage. 3. C(4, 2) and C(0, 0) 4. x = 4, y = 2, x + a =0, y + b = 0 5. a = -4; b = -2; (x, y) → (x - 4, y - 2) 6. A(-1, 4), B(1, 3), 7. y A B A B D C D 2 O C 2 x 4 2 8. yes 9. (x, y) → (x + 1, y - 3), A9(4, 3), B9(6, 2), D9(3, 0) Practice 9-2 1. (-3, -2) 2. (-2, -3) 3. (-1, -4) 4. (4, -2) 5. (4, -1) 6. (3, -4) y 7a. 4 2 6 4 2 O 2 I 4 L J 6 x K D(-2, 1) 84 Geometry All-In-One AnswersVersion A © Pearson Education, Inc., publishing as Pearson Prentice Hall. Chapter 9 All rights reserved. 8F: Vocabulary Review Puzzle Geometry: All-In-One Answers Version A (continued) y 7b. 12. 4 K I J 10 O 2 6 L y B A 2 4 6 x 6 4 T, T 4 2 y 8a. 4 Z 2 Y 6 4 2 All rights reserved. 6 4 2 2 2 4 W 4 8b. A6x X X A 4 B 1. a point at the origin, and two reflection lines 2. A reflection is an isometry in which a figure and its image have opposite orientations. 3. the image after two successive W reflections 6x 4 Z 4. y y=3 4 9. 6x Guided Problem Solving 9-2 4 Y 2 4 13. (-6, 4) 14. (-8, 0) 15. (0, -12) y 6 4 2 x y T 4 BO(0,0) 4 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2 6 4 2 2 2 6 x 4 5. y=3 4 B T A 10. T y 4 6 4 2 x O(0,0) T 2 4 4 A B B y O'(0,6) 6 x 6. y=3 4 y O'(0,6) A x 11. y T 4 O(0,0) T 4 O"(0,6) 8 6 4 2 A 2 B A 4 2 4 6 8 x 7. (0, - 6) 8. Yes. The x-coordinate remains 0 throughout. 9. O(0, 0)and O(0, 6) B All-In-One Answers Version A Geometry 85 Geometry: All-In-One Answers Version A Practice 9-3 (continued) 3. slope of OA 5 2 5 y 1. I 2. I 3. I 4. GH 5. G 6. ST 7. A U T x P O T S U S 4. y B 8. A O 9. x O P P N C Q O 5. slope of OB 5 252 ; slope of OC 5 52 ; slope of OD 5 252 ; N O P NQ M N L L Q O L M N L O P Q M P the slopes of perpendicular line segments are negative reciprocals. 6. square 7. yes 8. B(2, 7), C(7, -2), D(-2, -7) Practice 9-4 N L E 1. The helmet has reflectional symmetry. 2. The teapot has reflectional symmetry. 3. The hat has both rotational and reflectionalsymmetry. 4. The hairbrush has reflectional M O symmetry. Q Q 5. P P N O 11. L 12. 13. Q' P' S 6. M R' P P' S Q' 7. This figure has no lines of symmetry. 8. P R' Q Q line symmetry and 72° rotational symmetry 9. R R Guided Problem Solving 9-3 line symmetry 1. The coordinates of point A, and threerotation transformations. It is assumed that the rotations are counterclockwise. 2. parallelogram, rhombus, square 86 Geometry All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 10. D Q All rights reserved. P Geometry: All-In-One Answers Version A(continued) 10. 21. line symmetry and 90° rotational symmetry 11. Guided Problem Solving 9-4 1. the coordinates of one vertex of a figure that is symmetric about the y-axis 2. Line symmetry is the type of symmetry line symmetry and 45° rotational symmetry All rights reserved. 12. X for which there is areflection that maps a figure onto itself. 3. the coordinates of another vertex of the figure 4. images (and preimages) 5. reflection across the y-axis 6. (-3, 4) 7. Yes 8. (-6, 7) O X 180° rotational symmetry 13. line symmetry 14. Practice 9-5 1. L(-2, -2), M(-1, 0), N(2, -1), O(0, -1) 2. L(-30, -30), M(-15, 0), N(30,-15), O(0, -15) 3. L(-12, -12), M(-6, 0), N(12, -6), O(0, -6) 4. 53 5. 12 6. 2 7. yes 8. no 9. no R 10. O T A line symmetry and 45° rotational symmetry A R T 15. A 12. © Pearson Education, Inc., publishing as Pearson Prentice Hall. R 11. R A T R 180° rotational symmetry R 16. O A line symmetry 17. 19. 18.COOK 20. H O A X All-In-One Answers Version A T A T 13. P(-12, -12), Q(-6, 0), R(0, -6) 14. P(-12, 14), Q(1 14, -14), R(1 34 , 2) 15. P(-21, 6), Q(3, 24), R(-6, 6) 16. P(-2, 1), Q(-1, 0), R(0, 1) Guided Problem Solving 9-5 1. A description of a square projected onto a screen by an overhead projector,including the square’s area and the scale factor in relation to the square on the transparency. 2. The scale factor of a dilation is the number that describes the size change from an original figure to its image. 3. the area of the square on the transparency 4. smaller; The scale factor 16 > 1, so the dilation isan enlargement. 5. 1 ; 1 16 16 Geometry 87 Geometry: All-In-One Answers Version A y 8. 4 tranparency has 1 3 1 5 1 times the area. 7. 3 ft2 16 16 256 256 8. The shape does not matter. Regardless of the shape, the figure is being enlarged by a factor of 16 in two directions, so that the screen imagehas 16 16 = 256 as large an area as the figure on the transparency. 9. 2520 ft2 2 2 O 2 9. m m B y T S 4 2 O B 2 E 2 y 4 B C C C 6 4 2 O 2 E 4 5. y m J S 4 y 4 S 2 4 T 2 O 2 T y T 8 4 6 x B y 13. 4 T 2 4 S 2 4 2 O 2 4 B 6 E 2 6 x B 10 T 12. E E 4 x 4 m S B 6 E 2 O 2 y B S 2 J 4 x 2 11. J 7. T 2 m O 2 6x 4 10. 4. 2 8 x 6 T 2 B B B 6. 4 4 1. I. D II. C III. B IV. A 2. I. B II. A III. C IV. D 3. 2 2 4 x 4 2 O 2 4 E 2 4 6 x S 14. reflection 15. rotation 16. glide reflection 17. translation 88 Geometry All rights reserved. Practice 9-6 E S All-In-One Answers Version A © Pearson Education, Inc., publishing as PearsonPrentice Hall. 1 ;Being 1 as high and 1 as wide, the square on the 6. 256 16 16 (continued) Geometry: All-In-One Answers Version A (continued) Guided Problem Solving 9-6 7. 8. 1. assorted triangles and a set of coordinate axes 2. a transformation 3. the transformation that maps one triangle ontoanother 4. They are congruent, which confirms All rights reserved. that they could be related by one of the isometries listed. Corresponding vertices: E and P, D and Q, C and M. 5. counterclockwise; clockwise. The triangles have opposite orientations, so the isometry must be a reflection or glide reflection.6. No; there is no reflection line that will work. 7. a glide reflection consisting of a translation (x, y) → (x + 11, y) followed by a reflection across line y = 0 (the x-axis) 8. Yes, both are horizontal. 9. 180º rotation with center Q 22 1 , 0 R 2 11. Practice 9-7 1. 12. yes 13. yes 14. no 15. yes 16. no 17. notranslational symmetry 2. line symmetry, rotational symmetry, translational symmetry, glide reflectional symmetry © Pearson Education, Inc., publishing as Pearson Prentice Hall. 9.–11. Samples: 9. 10. Guided Problem Solving 9-7 1. two different-sized equilateral triangles and a trapezoid 2. A tessellationis a repeating pattern of figures that completely covers a plane, without gaps or overlaps. 3. Theorem 9-7 says that every quadrilateral tessellates. If the three polygons can be joined to form a quadrilateral (using each polygon at least once), that quadrilateral can be the basis of a tessellation. 4. 3. 5.Possible answer: translational symmetry 4. line symmetry, rotational symmetry, translational symmetry, glide reflectional symmetry 6. Check students’ work. 7. Answers may vary. Sample: 5. line symmetry, rotational symmetry, translational symmetry, glide reflectional symmetry 6. 9A: Graphic Organizerrotational symmetry, translational symmetry All-In-One Answers Version A 1. Transformations 2. Answers may vary. Sample: reflections; translations; compositions of reflections; and symmetry 3. Check students’ work. Geometry 89 Geometry: All-In-One Answers Version A (continued) 9B: ReadingComprehension 9E: Vocabulary Check 1. D 2. E, R, 5, and 20 3. The center of the county is Transformation: A change in the position, size, or shape of a geometric figure. Preimage: The given figure before a transformation is applied. Image: The resulting figure after a transformation is applied. Isometry:A transformation in which an original figure and its image are congruent. Composition: A transformation in which a second transformation is performed on the image of the first transformation. the center of the square mile bounded by roads 12, 13, K and L. 4. C 5. 10 miles 6. a 9C: Reading/Writing MathSymbols 1. reflection; opposite 2. rotation; same 3. glide reflection; opposite 4. translation; same 5. rotation; same 6. translation; same 9D: Visual Vocabulary Practice/High-Use Academic Words 1 2 P O All rights reserved. 9F: Vocabulary Review Puzzle 1. theorem 2. symbols 3. reduce 4. application 5.simplify 6. explain 7. characteristic 8. table 9. investigate G L Y H E T R Y T R Y D R O N I D 3 S Y M M E R O B 5 T U S E F S 6 I S 10 M E D L O M E E C L T R L I O A O 13 T 15 E T A I T O I A N R A 7 8 11 N S L R L O C U I O N N E 16 C 9 A T L R T A T N 14 I O R 12 D O I L A T I O N N U F D NE R 18 S 17 C A V L E N T Chapter 10 Practice 10-1 1. 8 2. 8 3. 60 4. 240 5. 2.635 6. 1.92 7. 0.8125 8. 9 9. 1500 10. 8.75 11. 3.44 12. 7 13. 110 14. 30 15. 16 16. 119 17. 12 18. 48 19. 40 20. 56 Geometry I T S O T M 90 C H E R I T O E N X Guided Problem Solving 10-1 1. 21 bh 2. Check students’work; B(7, 7) 3. a right angle å 4. 7 5. OA 6. 7 7. 24.5 8. a square 9. 49 10. 12 11. 24.5 12. 12 b2 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4 Geometry: All-In-One Answers Version A (continued) Practice 10-2 Practice 10-6 1. 48 cm2 2. 784 in.2 3. 11.4ft2 4. 90 m2 5. 2400 in.2 6. 374 ft2 7. 160 cm2 8. 176.25 in.2 9. 54 ft2 10. 42.5 square units 11. 96 " 3 square units 12. 36 " 3 square units 13. 45 cm2 14. 226.2 in.2 15. 49,500 m2 1. 32p 2. 16p 3. 7.8p 4. Samples: DFA, ABF 00 1 1 5. Samples: FE , FD 6. Samples: FEB , EDA 7. Samples: 0 0 FE , ED 8.ACB 9. FCD 10. 32 11. 43 12. 54 13. 39 14. 71 15. 90 16. 50 17. 220 18. 140 19. 220 20. 130 21. 6p in. 22. 16p cm 23. 94 p m Guided Problem Solving 10-2 1. 12 h(b1 1 b2) 2. 3; 4; 2 3. 9 4. no 5. yes 6. 3 All rights reserved. Practice 10-3 8 "3 8 "3 = 3 ;d = 3 1. x = 7; y = 3.5 " 3 2. a = 60; c 3. p = 4 " 3; q =8; x = 30 4. m1 = 45; m2 = 45; m3 = 90; m4 = 90 5. m5 = 60; m6 = 30; m7 = 120 6. m8 = 60; m9 = 30; m10 = 60 7. 12 " 3 8. 36.75 " 3 9. 75 " 3 10. 120 in.2 11. 137 in.2 12. 97 in.2 Guided Problem Solving 10-3 1. 10 2. The angles are all congruent and the sides are all congruent. 3. 360 4. 36 5. bisects 6. 27. 18 8. 180 9. 72 10. 180 11. 30; 15; 75 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Practice 10-4 1. 4 : 5; 16 : 25 2. 5 : 3; 25 : 9 3. 3 : 4; 9 : 16 4. 1 : 2 2700 2 2 5. 9 : 5 6. 1 : 6 7. 8 : 3 8. 576 5 in. 9. 7 cm 1152 2 10. 25 in. 11. 2 : 3 12. 1 : 2 13. 2 : 5 14. 108 ft2 Guided Problem Solving10-4 1. scale 2. Check students’ work. 3. Answers will vary. Let a be the side opposite the 30° angle, b be the side opposite the 50° angle, and c be the side corresponding to 200 yd. Then 200 yd 5 yd a < 20 mm, b < 30 mm and c < 40 mm. 4. 5 40 mm 1 mm 5. < 90 mm 6. < 16 mm 7. < 320 mm2 8. 450 9.8000 10. yes 11. 16,000 Practice 10-5 1. 174.8 cm2 2. 578 ft2 3. 1250.5 mm2 4. 192.6 m2 5. 1131.4 in.2 6. 419.2 cm2 7. 324.9 in.2 8. 162 cm2 9. 37.6 m2 10. 30.1 mi2 11. 55.4 km2 12. 357.6 in.2 13. 9.7 mm2 14. 384.0 in.2 15. 54.5 cm2 16. 9.1 ft2 17. 80.9 m2 18. 83.1 m2 19. 65.0 ft2 20. 119.3 ft2 21. 8m2 22. 55.4 ft2 11 Guided Problem Solving 10-6 1. 11; 2 2. 11; 3; 33 3. circumference; 2 4. 31 5. 26 6. 41 7. 105 8. 102 9. 3027.15 Practice 10-7 p 49 1 1 1. 49p 2. 21.2 3. 49 6 p 4. 6 p - 21.2 5. 4 p or 4 6. 8 p p 9 3 1 7. 16 p or 16 8. 16 - 81 9. 4p 10. 18 5 p 11. 8 p 12. 2 p p 63 25 13. 15 2 p 14. 6p 15. 8 p16. 3 p 17. 2 18. 3 19. 2 20. 10 Guided Problem Solving 10-7 ? pr 2 2. 45 3. 25 4. 30 5. 27 6. a piece from 1. mAB 360 the outer ring 7. yes 8. a piece from the top tier Practice 10-8 3 7 1 1. 8.7% 2. 10.9% 3. 15.3% 4. 10 5. 10 6. 25 7. 10 1 8. 1 9. 5 10. 22.2% 11. 4.0% 12. 37.5% 13. 20% 14. 50% 15. 40%16. 33 13 % Guided Problem Solving 10-8 1. outcomes 2. event 3. AF 4. FG 5. EG 6. AE; Check students’ work. 7. 2 or about 67% 8. 2 or about 3 3 67%; yes 9. 3 or 60% 5 10A: Graphic Organizer 1. Area 2. Answers may vary. Sample: areas of parallelograms and triangles; areas of trapezoids,rhombuses, and kites; areas of regular polygons; perimeters and areas of similar figures; trigonometry and area; circles and arcs; areas of circles and sectors; geometric probability 3. Check students’ work. 10B: Reading Comprehension 1. E 2. J 3. B 4. G 5. A 6. I 7. D 8. C 9. H 10. F 0 å å 11. a2 + b2 =c2 12. mAC 5 1058 13. MN RS 0 0 14. "225 5 15 15. #JKL 16. XY > PQ Guided Problem Solving 10-5 17. ~ABCD 18. 42 = 16 19. b 1. subtract 2. 50 3. 12 ap 4. 30 5. 4 6. tan4308 7. 166 8. 116 9. yes 10. 151 10C: Reading/Writing Math Symbols 1. 2860 ft2 2. 3331 ft2 3. 86 bundles 4. 2823 squares 5.$9.00 6. $27.00 7. 45 minutes 8. 21.5 hours 9. $645 10. $1419 All-In-One Answers Version A Geometry 91 (continued) 10D: Visual Vocabulary Practice Guided Problem Solving 11-2 1. area of a triangle 2. radius 3. altitude of a parallelogram 4. area of a trapezoid 5. area of a kite or rhombus 6. similarityratio 7. apothem 8. perimeter (of an octagon) 9. height of a triangle 1. a picture of a box, with dimensions 2. the amount of cardboard the box is made of, in square inches 3. A front and a back, each 4 in. 7 1 in.; a top and a bottom, each 2 4 in. 1 in.; and one narrow side, 1 in. 7 1 in. 4. Front and 2 back:each 30 in.2; top and bottom: each 4 in.2; one narrow side: 7 1 in.2 5. 75 1 in.2 6. 75 1 in.2 is about half a square 2 2 2 foot (1 ft2 = 144 in.2), which is reasonable. The answer is an approximation because the solution did not consider the overlap of cardboard at the glued joints, nor does it factor in thetrapezoid-shaped cutout in the front and back sides. 7. 3 21 in.2 Semicircle: Half a circle. Adjacent arcs: Arcs on the same circle with exactly one point in common. Apothem: The distance from the center to the side of a regular polygon. Circumference: The distance around a circle. Concentric circles:Circles that lie on the same plane and have the same center. 10F: Vocabulary Review Puzzle 1. adjacent arcs 2. semicircle 3. center 4. arc length 5. base 6. circle 7. central angle 8. congruent arcs 9. diameter 10. major arc 11. concentric circles 12. minor arc 13. Pythagorean triple 14. radius 15. apothem16. sector 17. height 18. circumference Highness, there is no royal road to geometry. Chapter 11 Practice 11-1 1. 8 2. 5 3. 12 4. C 5. D 6. B 7. A 8. triangle 9. rectangle 10. rectangle 11. rectangle Practice 11-3 1. 64 m2 2. 620 cm2 3. 188 ft2 4. 36p cm2 5. 800p m2 6. 300p in.2 7. 109.6 m2 8. 160.0 cm2 9.387.7 ft2 10. 147.6 m2 11. 118.4 cm2 12. 668.2 ft2 Guided Problem Solving 11-3 1. a solid consisting of a cylinder and a cone 2. the surface area of the solid 3. The top of the solid is a circular base of the cylinder, A1 = pr2. The side of the solid is the lateral area of the cylinder, A2 = 2prh. The bottom ofthe solid is the lateral area of the cone, A3 = pr/. 4. r = 5 ft; h = 6 ft; / 5 "52 1 122 5 13 ft 5. Top: A1 = 78.5 ft2; side: A2 = 188.5 ft2; bottom: 204.2 ft2 6. 471 ft2 7. About 0.43, or a little less than half. This seems reasonable. 8. 628 ft2 Practice 11-4 12. rectangle Guided Problem Solving 11-1 1. a two-dimensional network 2. Verify Euler’s Formula for the network shown. 3. 4 4. 6; 9 5. 4 + 6 = 9 + 1; This is true. 6. F becomes 3, V stays at 6, E becomes 8. Euler’s Formula becomes 3 + 6 = 8 + 1, which is still true. 7. 4 + 4 = 6 + 2; This is true. Practice 11-2 1. 62.8 cm2 2. 552.9 ft2 3. 226.2 cm2 4. 113.1m2 5. 44.0 ft2 6. 84.8 cm2 7a. 320 m2 7b. 440 m2 8a. 576 in.2 8b. 684 in.2 9a. 216 mm2 9b. 264 mm2 10a. 48 cm2 10b. 60 cm2 11a. 1500 m2 11b. 1800 m2 12a. 2000 ft2 12b. 2120 ft2 13. 8p m2 14. 94.5p cm2 15. 290p ft2 92 Geometry 1. 1131.0 m3 5. 785.4 cm3 10. 1728 ft3 14. 943 in.3 2. 94,247.8cm3 3. 7.1 in.3 4. 1781.3 in.3 6. 603.2 cm3 7. 120 in.3 8. 35 ft3 9. 42 m3 11. 784 cm3 12. 265 in.3 13. 76 ft3 15. 1152 m3 Guided Problem Solving 11-4 1. dimensions for a layer of topsoil, and two options for buying the soil 2. which purchasing option is cheaper 3. 1 ft 3 4. 1400 ft3 5. 467 bags; $1167.506. 1400 ft3 < 52 yd3 7. $1164 (or slightly less if the topsoil volume is not rounded up to the next whole number) 8. buying in bulk 9. The two costs are quite close. It makes sense that buying in bulk would be slightly cheaper for a good-sized backyard. 10. Buying topsoil by the bag would be less expensivebecause it would cost $335 and in bulk would cost at least $345.93. Practice 11-5 1. 34,992 cm3 2. 400 in.3 3. 10,240 in.3 4. 4800 yd3 5. 150 m3 6. 48 cm3 7. 628.3 cm3 8. 314.2 in.3 9. 4955.3 m3 10. 1415.8 in.3 11. 1005.3 m3 12. 18.8 ft3 13. 20 14. 2 15. 6 All-In-One Answers Version A © PearsonEducation, Inc., publishing as Pearson Prentice Hall. 10E: Vocabulary Check All rights reserved. Geometry: All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued) Guided Problem Solving 11-5 1. a description of a plumb bob, with dimensions 2. the plumb bob’s volume 3.equilateral; 2 cm 4. a = "3 cm; B = 6"3 cm2 5. Prism: V1 = 36"3 cm3; pyramid: V2 = 6"3 cm3; plum bob: V < 73 cm3 6. 1 ; this seems reasonable 7. 42 cm3 7 Practice 11-6 All rights reserved. 1. 2463.0 in.2 2. 6,157,521.6 m2 3. 12.6 cm2 4. 1256.6 m2 5. 50.3 ft2 6. 153.9 m2 7. 1436.8 mi3 8. 268,082.6 cm39. 7238.2 m3 10. 14.1 cm3 11. 2,572,354.6 cm3 12. 904,810.7 yd3 13. 546 ft2 14. 399 m2 15. 1318 cm2 16. 233 cm3 17. 7124 cm3 18. 5547 cm3 19. 42 cm3 Guided Problem Solving 11-6 1. the diameter of a hailstone, along with its density compared to normal ice 2. the hailstone’s weight 3. V 5 4 pr3 3 4.2.8 in. 5. 91.95 in.3 6. 1.7 lb 7. The density of normal ice. This was given for comparison purposes only. 8. 0.01 oz Practice 11-7 1. 7 : 9 2. 5 : 8 3. yes; 4 : 7 4. yes; 4 : 3 5. not similar 6. yes; 4 : 5 7. 125 cm3 8. 256 in.3 9. 432 ft3 10. 25 ft2 11. 90 m2 12. 600 cm2 13. 54,000 lb 14. 16 lb © PearsonEducation, Inc., publishing as Pearson Prentice Hall. Guided Problem Solving 11-7 1. two different sizes for an image on a balloon, and the volume of air in the balloon for one of those sizes 2. the volume of air for the other size 3. The two sizes of the clown face have a ratio of 4 : 8, or 1 : 2. 4. 1 : 8 5. 108in.3 8 = 864 in.3 6. Yes, the two clown face heights are lengths, measured in inches (not in.2 or in.3). 7. 351 cm3 11A: Graphic Organizer 1. Surface Area and Volume 2. Answers may vary. Sample: space figures and nets; space figures and drawing; surface areas; and volumes. 3. Check students’ work.11B: Reading Comprehension 1. Area = 12 base times height; area of a triangle 2. slope = the difference of the y-coordinates divided by the difference of the x-coordinates; slope of the line between 2 points 3. a squared plus b squared equals c squared, the Pythagorean Theorem; the lengths of the sidesof a right triangle 4. area = pi times radius squared; area of a circle 5. volume = pi times radius squared times height; volume of a cylinder 6. the sum of the x-coordinates of two points divided by 2, and the sum of the y-coordinates of two points divided by 2; midpoint of a line segment with endpoints (x1,y1) and (x2, y2) 7. circumference = two times pi times the radius; circumference of a circle 8. volume = one-third pi times radius squared times height; volume of a cone All-In-One Answers Version A 9. distance equals the square root of the quantity of the sum of the square of the difference of the x-coordinates, squared, plus the difference of the y-coordinates, squared; the distance between two points (x1, y1) and (x2, y2) 10. area = one-half the height times the sum of base one plus base two; area of a trapezoid 11. a 11C: Reading/Writing Math Symbols 1. 527.8 in2 2. 136 ft2 11D: VisualVocabulary Practice 1. prism 2. cross section 3. sphere 4. cylinder 5. cone 6. pyramid 7. polyhedron 8. altitude 9. base 11E: Vocabulary Check Face: A surface of a polyhedron. Edge: An intersection of two faces of a polyhedron. Altitude: For a prism or cylinder, a perpendicular segment that joins theplanes of the bases. Surface area: For a prism, cylinder, pyramid, or cone, the sum of the lateral area and the areas of the bases. Volume: A measure of the space a figure occupies. 11F: Vocabulary Review 1. polyhedron 2. edge 3. vertex 4. cross section 5. prism 6. altitude 7. right prism 8. slant height 9.cone 10. volume 11. sphere 12. hemispheres 13. similar figures 14. hypotenuse Chapter 12 Practice 12-1 1. 32 2. 50 3. 72 4. 15 5. "91 6. 6 7. "634 8. "901 9. 4 "3 10. circumscribed 11. inscribed 12. circumscribed 13. 24 cm 14. 28 in. 15. 52 ft Guided Problem Solving 12-1 1. The area of one is twice thearea of the other. 2. inscribed; circumscribed 3. 4. r 5. 2r 6. 4r2 7. "2r 8. 2r 2 9. 4r 2 = 2(2r 2): One area is double the other area. 10. yes 11. The same: The area of one circle is twice the area of the other circle. Geometry 93 Geometry: All-In-One Answers Version A 0 0 3. r = #2 ; mAB 102.7 4. 3 5. 4.5 6. 37. 20.8 8. 13.7 9. 8.5 Q T; 010. 0 PR SU 11. A J; BC KL 12. Construct radii OD and OB . FD = EB because a diameter perpendic41 0 (0, 5) (0, 0) ular to a chord bisects it, and chords CD and AB are congruent (given). Then, by HL, OEB OFD, and by CPCTC, OE = OF. 13. Because congruent arcs havecongruent chords, AB = BC = CA. Then, because an equilateral triangle is equiangular, mABC = mBCA = mCAB. Guided Problem Solving 12-2 y 18. (3, 5) x 1 2 3 4 5 y 19. 2 2 2 4 x 2 (2, 4) 4 Guided Problem Solving 12-3 1. C 2. 180 3. a diameter 4. Check students’ work. 5. Check students’ work. 6.Check students’ work. 7. Check students’ work. 8. 90 9. Check students’ work. 1. 87 2. 35 3. 45 4. 120 5. x = 45; y = 50; 6. 90; y = 70 7. x = 58; y = 58; z = 63 8. x = 30; y = 66 9. x = 30; y = 30; z = 120 10. x = 16; y = 52 11. x = 138; y = 111; z = 111 12. x = 30; y = 60 13. 10 14. 14.8 15. 4.7 16. 4 17. 3.2 18.6 3 5 4 3 2 1 Practice 12-3 Practice 12-4 x 5 1. perpendicular 2. bisects 3. 4; 3 4. right 5. radius 6. Pythagorean Theorem 7. 5 8. yes 9. 8; 4 1. A and D; B and C 2. ADB and CDB 3. ADB and CAD 4. 55 5. x = 45; y = 50; z = 85 6. x = 90; y = 70 7. 180 8. 70 9. x = 120; y = 60; z = 60 10. x = 120; y = 100; z =140 11. x =0 63; y = 63; z = 54 12. x = 50; y = 80; z = 80 13a. mAE = 170 13b. mC = 85 13c. mBEC = 10 13d. mD = 85 14a. mA = 90 14b. mB = 80 14c. mC = 90 14d. mD = 100 (5, 0) All rights reserved. 1. r = 13; mAB 134.8 2. r = 3 "5; mAB 53.1 12. (x + 3)2 + (y - 3)2 = 1 13. x2 + (y - 3)2 = 16 14. (x - 7)2 +(y + 2)2 = 4 15. x2 + (y + 20)2 = 100 16. (x + 4)2 + (y + 6)2 = 25 y 17. 4 6 20. y (1, 7) 6 4 (1, 1) Guided Problem Solving 12-4 4 2 1. tangent 2. 360. 3. 12 4. 85 5. 2 6. 180 7. 95 8. 104; 86; 75 9. 360 10. yes; The sum of the measures of the arcs should be 360. 11. Opposite angles are not 6 2 (5, 1) 2 4 x 2 4supplementary. Practice 12-5 1. C(0, 0), r = 6 2. C(2, 7), r = 7 3. C(-1, -6), r = 4 4. C(-3, 11), r = 2"3 5. x2 + y2 = 49 6. (x - 4)2 + (y - 3)2 = 64 7. (x - 5)2 + (y - 3)2 = 4 8. (x + 5)2 + (y - 4)2 = 14 9. (x + 2)2 + (y + 5)2 = 2 10. (x + 1)2 + (y - 6)2 = 5 11. x2 + y2 = 4 94 Geometry 21. 23. 24. 25. 26. x2 (x (x (x (x + + +- y2 = 25 4)2 + (y 7)2 + (y 4)2 + (y 4)2 + (y 22. (x - 5)2 + (y - 9)2 = 9 + + + - 3)2 2)2 3)2 1)2 = = = = 61 80 4 16 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. Practice 12-2 (continued) Geometry: All-In-One Answers Version A (continued) 13. Guided ProblemSolving 12-5 1. (2, 2); 5 2. 43 3. They are perpendicular. 4. - 34 5. 39 4 6. y 5 234 x 1 39 7. 13 8. y 5 23 x 1 39 ; yes 4 4 4 9. x-intercept: a = 25 ; y-intercept: b = 25 3 4 0.5 cm T U 14. 5 mm 6 mm Practice 12-6 1. Guided Problem Solving 12-6 T 1. a line 2. Check students’ work. 3. They are perpendicular.4. - 1 5. 2 6. the midpoint of PQ 7. (3, 2) 2 All rights reserved. 1.5 cm 8. -4 9. y = 2x - 4 10. (2, 0); yes 11. y 5 21 x 1 2 2. 1 in. P 12A: Graphic Organizer 1. Circles 2. Answers may vary. Sample: tangent lines; chords Q and arcs; inscribed angles; and angle measures and segment lengths 3. Checkstudents’ work. 3. 12B: Reading Comprehension A B 4. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Answers may vary. Sample answers: 1. AD, CE 2. BF, CD, AD 3. HD 4. AH 1 1 1 0 00 0 0 5. AF , AB , BC 6. CDF, DAB, DAC 7. AC , ED 1 8. /AOC 9. /DCE 10. DEA 11. /DAH 12. /AGB13. /AHC 14. /DCE 15. AH 16. a 0.75 in. 0.5 in. R 0.5 in. S 12C: Math Symbols * )Reading/Writing * ) 1. BC ' DE 2. EF is tangent to (G. 3. WX > ZX 4. AB ' XY 5. HM = 4 0.75 in. 5. 6. 12D: Visual Vocabulary Practice Y X B A 7. 1. tangent to a circle 2. standard form of an equation of a circle 3. secant 4.chord 5. circumscribed about 6. intercepted arc 7. inscribed in 8. inscribed angle 9. locus H F E P J C Inscribed in: A circle inside a polygon with the sides of the G polygon tangent to the circle. 8. a third line parallel to the two given lines and midway between them 9. a sphere whose center is the givenpoint and whose radius is the given distance 10. the points within, but not on, a circle of radius 1 in. centered at the given point 11. 12E: Vocabulary Check 12. Chord: A segment whose endpoints are on a circle. Point of tangency: The single point of intersection of a tangent line and a circle. Tangent to acircle: A line, segment, or ray in the plane of a circle that intersects the circle in exactly one point. Circumscribed about: circle outside a polygon with the vertices of the polygon on the circle. 0.75 in. 2 cm Q All-In-One Answers Version A R S 0.75 in. 12F: Vocabulary Review 1. D 2. F 3. G 4. A 5. C 6. B 7. E8. M 9. I 10. L 11. N 12. H 13. K 14. J Geometry 95 Geometry: All-In-One Answers Version A Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date ________________ Lesson 1-1 3 Finding a Counterexample Find a counterexample for the conjecture. Patterns and Inductive Reasoning Lesson Objective 1 Since 32 42 52, the sum of the squares of two consecutive numbers is the square of the nextconsecutive number. NAEP 2005 Strand: Geometry Use inductive reasoning to make conjectures Topic: Mathematical Reasoning Begin with 4. The next consecutive number is 4 1 5 . Local Standards: ____________________________________ The sum of the squares of these two consecutive numbersis 42 52 16 25 41 . Vocabulary. The conjecture is that 41 is the square of the next consecutive number. The square of the next consecutive number is (5 1)2 62 36 . Inductive reasoning is reasoning based on patterns you observe. A conjecture is a conclusion you reach using inductive reasoning. Acounterexample is an example for which the conjecture is incorrect. All rights reserved. All rights reserved. So, since 42 52 and 62 are not the same, this conjecture is Use the pattern to find the next two terms in the sequence. 384, 192, 96, 48, . . . half 2 48 2 2000, $9.50 in 2001, and $11.00 in 2002.Make a conjecture about the price in 2003. Write the data in a table. Find a pattern. 24 and 24 2 12 . 2 Using Inductive Reasoning Make a conjecture about the sum of the cubes All rights reserved. of the first 25 counting numbers. Find the first few sums. Notice that each sum is a perfect square and thatthe perfect squares form a pattern. 3 1 3 1 3 1 3 1 3 1 2 3 2 3 3 2 3 3 3 3 2 3 4 3 3 3 3 2 3 4 5 1 1 2 9 3 2 36 6 2 100 10 2 225 15 2 1 2 (1 2) 2 (1 2 3) 2 (1 2 3 4) 2 (1 2 3 4 5) The sum of the first two cubes equals the square of the sum of the first two counting numbers. The sum of the first three cubesequals the square of the sum of the first three counting numbers. This pattern continues for the fourth and fifth rows. So a conjecture might be that 2000 2001 2002 $8.00 $9.50 $11.00 Each year the price increased by $ 1.50 . A possible conjecture is that the price in 2003 will increase by $ 1.50 . If so, theprice in 2003 would be $11.00 $ the preceding term. The next two © Pearson Education, Inc., publishing as Pearson Prentice Hall. terms are 48 96 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Each term is 192 2 . 4 Applying Conjectures to Business The price of overnight shippingwas $8.00 in Examples. 1 Finding and Using a Pattern Find a pattern for the sequence. 384 false 1.50 $ 12.50 . Quick Check. 1. Find the next two terms in each sequence. a. 1, 2, 4, 7, 11, 16, 22, 29 , 37 b. Monday, Tuesday, Wednesday, ,... Thursday , Friday ,... c. , ,... Answers may vary. Sample: the

sum of the cubes of the first 25 counting numbers equals the square of the sum of the first 25 counting numbers, or (1 2 3 . . . 25) 2. 2 Geometry Lesson 1-1 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ 1 12 Lesson 1-2 1 3 4 22 1 3 5 9 32 1 3 5 7 16 42 1 3 5 7 9 25 52 Lesson Objectives 1 Make isometric and orthographic drawings 2 Draw nets for three-dimensional figures Drawings, Nets, and Other Models NAEP 2005 Strand: Geometry Topic: Dimension and Shape LocalStandards: ____________________________________ Vocabulary. All rights reserved. All rights reserved. An isometric drawing of a three-dimensional object shows a corner view of the figure drawn The sum of the first 35 odd numbers is 35 2, or 1225. 3. Finding a Counterexample Find acounterexample for the conjecture. Some products of 5 and other numbers are shown in the table. 5 7 35 5 3 15 5 11 55 4. Suppose the price of two-day shipping was $6.00 in 2000, $7.00 in 2001, and $8.00 in 2002. Make a conjecture about the price in 2003. Each year the price increased by $1.00. Apossible conjecture is that the price in 2003 will increase by $1.00. If so, the price in 2003 would be $8.00 $1.00 $9.00. All-In-One Answers Version A Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 Isometric Drawing Make an isometric drawing of Since 5 2 10,not all products of 5 end in 5. (But they all either end in 5 or 0.) Geometry Lesson 1-1 three-dimensional figure. A foundation drawing shows the base of a structure and the height of each part. Examples. Therefore, the product of 5 and any positive integer ends in 5. 4 on isometric dot paper. Anorthographic drawing is the top view, front view, and right-side view of a A net is a two-dimensional pattern you can fold to form a three-dimensional figure. 5 13 65 5 9 45 5 25 125 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson PrenticeHall. 3 Name_____________________________________ Class____________________________ Date ________________ 2. Make a conjecture about the sum of the first 35 odd numbers. Use your calculator to verify your conjecture. 1 Geometry Lesson 1-1 the cube structure at right. An isometricdrawing shows three sides of a figure corner from a view. Hidden edges are not shown, so all edges are solid lines. Daily Notetaking Guide Geometry Lesson 1-2 Geometry 5 1 Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date ________________ 2 Orthographic Drawing Make an orthographic drawing of the 4 Identifying Solids from Nets Is the pattern a net for acube? isometric drawing at right. If so, name two letters that will be opposite faces. A Orthographic drawings flatten the depth of a figure. An orthographic three drawing shows views. Because no edge of the isometric drawing is hidden in the top, front, and right views, all lines are solid. The pattern a netbecause you can is fold it to form a cube. Fold squares A and C up to form the back and front of the cube. Fold D up to form a side. Fold E over to form the top. Fold F down to form another side. B Fr ht t Rig on D E F C After the net is folded, the following pairs of letters are on opposite faces: A and C arethe back and front faces. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Top Identify the square that represents the tallest part. Write the number 2 in the back square to indicate that the back section is 2 cubes high. Write the number 1 in each of the two front squares to indicate that eachfront section is 1 cube high. 1 1 Right 2 5 Drawing a Net Draw a net for the figure with a square base and four isosceles triangle faces. Label the net with its dimensions. Think of the sides of the square base as hinges, and “unfold” the figure at these edges to form a net. The base of each of the foursquare isosceles triangle faces is a side of the . Write in the known dimensions. ht t Rig on Because the top view is formed from 3 squares, show 3 squares in the foundation drawing. faces. top 8 cm 10 8 cm cm All rights reserved. Fr and cm isometric drawing. To make a foundation drawing, use the topview of the orthographic drawing. bottom D and F are the right and left side faces. 10 3 Foundation Drawing Make a foundation drawing for the All rights reserved. Right © Pearson Education, Inc., publishing as Pearson Prentice Hall. Top All rights reserved. B and E are the Front Quick Check. 1. Make anisometric drawing of the cube structure below. Front Geometry Lesson 1-2 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ Geometry Lesson 1-2Name_____________________________________ Class____________________________ Date ________________ 2. Make an orthographic drawing from this isometric drawing. 5. The drawing shows one possible net for the Graham Crackers box. 14 cm Fr ht on t Rig 20 cm Front Top 7 Right 7 cmAM AH GR KERS AC CR AM AH GR KERS AC CR © Pearson Education, Inc., publishing as Pearson Prentice Hall. 6 20 cm 7 cm 14 cm 3. a. How many cubes would you use to make the structure in Example 3? 4 All rights reserved. All rights reserved. Draw a different net for this box. Show thedimensions in your diagram. Answers may vary. Example: 14 cm 20 cm b. Critical Thinking Which drawing did you use to answer part (a), the foundation drawing or the isometric drawing? Explain. 7 cm 4. Sketch the three-dimensional figure that corresponds to the net. 8 2 Geometry Lesson 1-2 GeometryDaily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Answers may vary. Sample: the foundation drawing; you can just add the three numbers. Daily Notetaking Guide Geometry Lesson 1-2 9 All-In-OneAnswers Version A Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 1-3 1 A plane is a flat surface that has no thickness. Points, Lines, and Planes Lesson ObjectivesNAEP 2005 Strand: Geometry Understand basic terms of geometry Understand basic postulates of geometry 2 Name_____________________________________ Class____________________________ Date ________________ Two points or lines are coplanar if they lie on the same plane. Topic:Dimension and Shape Local Standards: ____________________________________ A postulate or axiom is an accepted statement of fact. Vocabulary and Key Concepts. Examples. t Line t is the only line that passes through points A and B . B A name three points that are collinear and three points thatare not collinear. All rights reserved. All rights reserved. Through any two points there is exactly one line. Points *AE ) and *BD ) intersect at C . Z Y W triangle and are not collinear. U R You can name a plane using any three or more points on that plane that are not collinear. Plane RST and plane STWintersect in ST . Postulate 1-4 Through any three noncollinear points there is exactly one plane. A point is a location. Space is the set of all points. A line is a series of points that extends in two opposite directions without end. Geometry Lesson 1-3 t B A Collinear points are points that lie on the same line.RSU plane RST , plane plane STU , and plane , plane T S Some possible names for the plane shown are the following: © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. *ST) R T W , RTU . RSTU H 3 Finding the Intersectionof Two Planes Use the diagram at right. What is the intersection of plane HGC and plane AED? G E F As you look at the cube, the front face is on plane AEFB, the back face is on plane HGC, and the left face is on plane AED. The back and left faces of the cube intersect at *HD) vertically at C D A .Planes HGC and AED intersect HD B . Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-3 11 Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________Class____________________________ Date________________ 4 Using Postulate 1-4 Shade the plane that contains X, Y, and Z. Lesson 1-4 Z Points X, Y, and Z are the vertices of one of the four triangular faces of the pyramid. To shade the plane, shade the interior of the triangle formed by X , Y , andZ . Y X V W Quick Check. Local Standards: ____________________________________ Segment AB A segment is the part of a line consisting of two AB b. Name line m in three different ways. )* ) Answers may vary. Sample: ZW , WY , YZ . All rights reserved. All rights reserved. endpoints and all pointsbetween them. no )* NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures Vocabulary. 1. Use the figure in Example 1. a. Are points W, Y, and X collinear? * Segments, Rays, Parallel Lines and Planes Lesson Objectives 1 Identify segments and rays 2 Recognize parallel lines B AEndpoint Endpoint ) A ray is the part of a line consisting of one endpoint Ray YX YX and all the points of the line on one side of the endpoint. X Y Endpoint c. Critical Thinking Why do you think * ) arrowheads are used when drawing a line or naming a line such as ZW ? Opposite rays are two collinear rayswith the same Arrowheads are used to show that the line extends in opposite directions without end. Q ) endpoint. RQ H G Z Answers may vary. Sample: HEF, HEFG, FGH. E * ) 3. Name two planes that intersect in BF . F H ABF and CBF E G F C D A 4. a. Shade plane VWX. Z B b. Name a point that iscoplanar with points V, W, and X. Y V 12 ) R RS S are opposite rays. Skew lines are noncoplanar; therefore, they are not parallel and do not intersect. C D H E ←→ AB is ←→ B A parallel ←→ AB and CG are ← → to EF. skew lines. G F Parallel planes are planes that do not intersect. G A J D Y andParallel lines are coplanar lines that do not intersect. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2. List three different names for plane Z. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. m and Z form a If two planes intersect, then they intersect inexactly one line. © Pearson Education, Inc., publishing as Pearson Prentice Hall. lie on a line, so they are 2 Naming a Plane Name the plane shown in two different ways. Postulate 1-3 10 W , and so no other set of three points is collinear. For example, X, Y, B E S Z , Any other set of three points in thefigure do not lie on a line, If two lines intersect, then they intersect in exactly one point. D Y collinear. Postulate 1-2 C X 1 Identifying Collinear Points In the figure at right, Postulate 1-1 A B C A Plane ABC X H B Plane ABCD is parallel to plane GHIJ. I C W Geometry Lesson 1-3 All-In-One Answers VersionA Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-4 Geometry 13 3 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________Name_____________________________________ Class____________________________ Date ________________ 2. Use the diagram in Example 2. a. Name all labeled segments that are parallel to GF. Examples. 1 Naming Segments and Rays Name the segments and rays in the figure. A HE, CB,DA The labeled points in the figure are A, B, and C. A segment is a part of a line consisting of two endpoints and all points between them. A segment is named by its two endpoints. So the BA (or AB) segments are and . BC (or CB) B A ray is a part of a line consisting of one endpoint and all the points ofthe line on one side of that endpoint. A ray is named by its endpoint first, followed ) . ) by any other point on the ray. So the rays are and BA BC C B A Parallel segments lie in the same plane, and the lines that contain them do not intersect. The three segments in the figure that are parallel to AE are BF ,CG and DH . H All rights reserved. c. Name another pair of parallel segments and another pair of skew segments. D Name all segments that are parallel to AE. Name all segments that are skew to AE. AB, DC, AE, DH All rights reserved. 2 Identifying Parallel and Skew Segments Use the figure at right. b.Name all labeled segments that are skew to GF. C Answers may vary. Sample: CG, BF; EF, DH 3. Use the diagram to the right. a. Name three pairs of parallel planes. G E S PSWT RQVU, PRUT SQVW, PSQR TWVU W F Skew segments are segments that do not lie in the same plane. The foursegments in the figure that do not lie in the same plane as AE are BC , GH . do not intersect Planes are parallel if they opposite . If the walls of your classroom walls are parts of parallel planes. If the ceiling and floor of the classroom are level, they are parts of parallel planes. Quick Check. ) ) 1. CriticalThinking Use the figure in Example 1. CB and BC form a line. Are they opposite rays? Explain. No; they do not have the same endpoint. 14 Geometry Lesson 1-4 Name_____________________________________ Class____________________________ Date________________ TV c. Name a line thatis parallel to plane QRUV. * ) Answers may vary. Sample: PS 15 Geometry Lesson 1-4 Name_____________________________________ Class____________________________ Date ________________ Measuring Segments Lesson Objectives 1 Find the lengths of segments * ) Daily NotetakingGuide Daily Notetaking Guide Lesson 1-5 * ) b. Name a line that is parallel to PQ . All rights reserved. 3 Identifying Parallel Planes Identify a pair of parallel planes in your classroom. are vertical, V U Examples. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes 1 Comparing SegmentLengths Find AB and BC. A Are AB and AC congruent? Local Standards: ____________________________________ BC = ∆ –1 Vocabulary and Key Concepts. B C 4 3 2 1 0 1 2 3 4 5 AB = ∆-5 - (–1) « = ∆-4« = 4 - 4« = »–5» = 5 AB BC so AB and AC are not congruent. . All rights reserved. The points ofa line can be put into one-to-one correspondence with the real numbers so that the distance between any two points is the absolute value of the difference of the corresponding numbers. All rights reserved. Postulate 1-5: Ruler Postulate Postulate 1-6: Segment Addition Postulate 2 Using the SegmentAddition Postulate If AB 25, find the value of x. Then find AN and NB. AN ( 2x 6 )( A AB ) 25 1 25 3x B is between A and C, then AB BC AC. NB x7 3x If three points A, B, and C are collinear and B C ( )6 NB x 7 ( 8 ) 7 R AB 5 u a Q b u A B a b a coordinate of A coordinate of B Congruent () segments aresegments with the same length. 2 cm A B A D C 2 cm C B AB CD AB CD D © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. the length of AB 24 x 8 AN 2x 6 2 8 A coordinate is a point’s distance and direction from zero on anumber line. 2x 6 x7 A N Use the Segment Addition Postulate (Postulate 1-6) to write an equation. B Segment Addition Postulate Substitute. Simplify the left side. Subtract 1 from each side. Divide each side by 3 . 10 15 Substitute 8 for x. AN 10 and NB 15 , which checks because the sum of the segmentlengths equals 25. midpoint A midpoint is a point that divides a segment into two congruent segments. A | AB 16 4 Geometry Lesson 1-5 Geometry B | C BC Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-5 17 All-In-One Answers Version A © Pearson Education, Inc., publishing asPearson Prentice Hall. and © Pearson Education, Inc., publishing as Pearson Prentice Hall. FG T © Pearson Education, Inc., publishing as Pearson Prentice Hall. CD , Q R P Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date ________________ R Find RM, MT, and RT. Name_____________________________________ Class____________________________ Date ________________ 2. EG 100. Find the value of x. Then find EF and FG. 8x 36 5x 9 3 Using the Midpoint M isthe midpoint of RT. M 4x – 20 T E Use the definition of midpoint to write an equation. RM 8x 36 Substitute. 8x Add 45 3 x x ( 15 15 )9 ) 36 84 Substitute 168 , which is half of 84 15 for x. 84 RT RM MT RM and MT are each from each side. Divide each side by 3 . Segment Addition 168 Postulate , thelength of RT. All rights reserved. ( RM 5x 9 5 MT 8x 36 8 to each side. 5x © Pearson Education, Inc., publishing as Pearson Prentice Hall. 15 36 Subtract x 15, EF 40; FG 60 All rights reserved. 45 G midpoint Definition of © Pearson Education, Inc., publishing as Pearson Prentice Hall. 5x 9 5x MT 2x + 30F 3. Z is the midpoint of XY, and XY 27. Find XZ. 13.5 Quick Check. 1. Find AB. Find C, different from A, such that AB and AC are congruent. A B 7 6 5 4 3 2 1 0 1 2 3 4 5 AB |7 (2)| |5| 5 We also want BC 5. That means that C must lie 5 units on Geometry Lesson 1-5 Daily Notetaking Guide DailyNotetaking Guide Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ An angle () is formed by two rays with thesame endpoint. The rays Measuring Angles Lesson Objectives 1 Find the measures of angles 2 Identify special angle pairs TBQ x° A straight angle has a measurement of exactly 180. Congruent angles are two angles with the same measure. Examples. O C If AOC is a straight angle, then mAOB mBOC180 . B Classify each as acute, right, obtuse, or straight. a. 80 17 0 70 12 0 60 13 0 50 y x 10 A 80 C 0 17 0 Daily Notetaking Guide B 10 0 1 10 20 180 90 1 16 0 10 0 80 7 0 10 0 6 0 11 0 20 50 0 1 3 4 y x 17 0 All-In-One Answers Version A 13 0 50 20 20 60 16 0 10 90 1 0 15 0 30 14 0 7 0 10 0 6 0 110 20 50 0 1 3 10 0 1 10 80 12 70 0 60, acute Daily Notetaking Guide A b. C O Geometry Lesson 1-6 3 G 2 Measuring and Classifying Angles Find the measure of each /AOC. A 20 . 4 A O C lG Finally, the name can be a point on one side, the vertex, and a point on the other side of the angle: lAGC orlCGA . 40 B The name can be the vertex of the angle: . 0 A mAOC. 14 BOC 30 m 15 0 If point B is in the interior of AOC, then C l3 The name can be the number between the sides of the angle: 0 15 0 30 14 0 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Postulate 1-8: Angle AdditionPostulate 1 Naming Angles Name the angle at right in four ways. 20 B 16 0 10 17 0 180 O AOB A right angle has a measurement of exactly 90. 10 30 16 0 15 0 y 0 m x 5 180 180 An obtuse angle has a measurement between 90 and 180. 13 0 50 x straight angle angle ,x, 17 0 60 1 D 40 10 0 1 10 80 1270 0 14 90 17 0 20 10 0 20 10 80 obtuse 90 90 An acute angle has a measurement between 0 and 90. 30 20 50 0 1 3 11 0 x5 16 0 A 70 angle right 90 15 0 60 14 0 40 C . mlCOD 5 u x 2 y u . All rights reserved. 180 b. If OC is paired with x and OD is paired with y, then © Pearson Education, Inc.,publishing as Pearson Prentice Hall. ) ) and OB is paired with ) angle 0,x, x° x° x° acute All rights reserved. Postulate 1-7: Protractor Postulate ) ) ) ) Let OA and OB be opposite rays in a plane. OA , OB , and all the rays with * ) endpoint O that can be drawn on one side of AB can be paired with the realnumbers from 0 to 180 so that a. OA is paired with 0 T 1 Q Local Standards: ____________________________________ Vocabulary and Key Concepts. ) B are the sides of the angle and the endpoint is the vertex of the angle. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes 0 150 30 14 0 4 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson 1-6 19 Geometry Lesson 1-5 0 18 16 0 All rights reserved. the other side of B from A. So C must lie at –2 5 3. 180 B O 150, obtuse Geometry Lesson 1-6 Geometry 21 5 Geometry: All-In-One Answers Version A(continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date ________________ 3 Using the Angle Addition Postulate Suppose thatm1 42 and mABC 88. Find m2. Quick Check. A 1. a. Name CED two other ways. Use the Angle Addition Postulate (Postulate 1-8) to solve. m 1 m 2 mABC 42 Angle Addition m2 88 Substitute 42 for mABC. m2 46 Subtract 42 l2, lDEC Postulate for m1 and 1 88 2 B from each side. b. Critical ThinkingWould it be correct to name any of the angles E? Explain. C No, 3 angles have E for a vertex, so you need more information in the name to distinguish them from one another. ⬔ 3 and ⬔ 4 1 All rights reserved. numbered angles that are related as follows: a. complementary 2 3 5 b. supplementary Allrights reserved. 4 Identifying Angle Pairs In the diagram identify pairs of 4 ⬔ 1 and ⬔ 2; ⬔ 2 and ⬔ 3 2. Find the measure of the angle. Classify it as acute, right, obtuse, or straight. 60; acute c. vertical angles ⬔ 1 and ⬔ 3 G 3. If mDEG 145, find mGEF. Yes; the markings show they are congruent. B C Db. B and ACD are supplementary No; there are no markings. c. m(BCA) m(DCA) 180 Yes; you can conclude that the angles are supplementary from the diagram. E F A E D a. ⬔ AOB b. ⬔ BOC or ⬔ DOE O B C 5. Can you make each conclusion from the information in the diagram? Explain. a. 1 3 b. 4and 5 are supplementary c. m(1) m(5) 180 5 1 4 2 3 a. Yes; the markings show they are congruent. b. Yes; you can conclude the angles are adjacent and supplementary from the diagram. c. Yes; you can conclude that the angles are supplementary from the diagram and their sum is 180 by the AngleAddition Postulate. d. AB BC Yes; the markings show they are congruent. 22 D 4. Name an angle or angles in the diagram supplementary to each of the following: a. DOA b. EOB All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. A conclusion from the diagram? a. A C ©Pearson Education, Inc., publishing as Pearson Prentice Hall. 35 5 Making Conclusions From a Diagram Can you make each Geometry Lesson 1-6 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-6 23 Name_____________________________________Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 1-7 2 Constructing Congruent Angles Construct Y so that Y G. Basic Constructions Step 1 Draw a ray withendpoint Y. NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures Y G Local Standards: ____________________________________ 75 Step 2 With the compass point on point G, draw an arc that intersects both sides of G. Label the points of intersection E and F. Vocabulary. ©Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson Objectives 1 Use a compass and a straightedge to construct congruent segments and congruent angles 2 Use a compass and a straightedge to bisect segments and angles E A straightedge is a ruler with no markings on it. A compassis a geometric tool used to draw circles and parts of circles called arcs. Perpendicular lines are two lines that intersect to form right angles. A perpendicular bisector of a segment is a line, segment, or ray that is A D C B All rights reserved. All rights reserved. Construction is using a straightedge and acompass to draw a geometric figure. G 75 F Step 3 With the same compass setting, put the compass point on point Y. Draw an arc that intersects the ray. Label the point of intersection Z. perpendicular to the segment at its midpoint, thereby bisecting the segment K N L Examples. 1 ConstructingCongruent Segments Construct TW congruent K to KM. Step 1 Draw a ray with endpoint T. M T Step 2 Open the compass the length of KM. Step 3 With the same compass setting, put the compass point on point T. Draw an arc that intersects the ray. Label the point of intersection W. T W © PearsonEducation, Inc., publishing as Pearson Prentice Hall. J coplanar angles. © Pearson Education, Inc., publishing as Pearson Prentice Hall. into two congruent segments. An angle bisector is a ray that divides an angle into two congruent Z Y Step 4 Open the compass to the length EF. Keeping the samecompass setting, put the compass point on Z. Draw an arc that intersects with the arc you drew in Step 3. Label the point of intersection X. X Y ) Step 5 Draw YX to complete Y. Z X KM TW Y G 24 6 Geometry Lesson 1-7 Geometry Daily Notetaking Guide Daily Notetaking Guide Y Z Geometry Lesson 1-725 All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________Date ________________ 3 Constructing the Perpendicular Bisector A Given: AB. * ) * ) Construct: XY so that XY AB at the midpoint M of AB. Quick Check. B 1. Use a straightedge to draw XY. Then construct RS so that RS 2XY. Step 1 Put the compass point on point A and draw a long arc. greater Besure that the opening is than 12AB. X A B * ) Step 3 Draw XY . The point of intersection of AB and XY is M, midpoint the of AB. M Y ) 4 Finding Angle Measures WR bisects AWB so that mAWR x R All rights reserved. 4x 48 B W m m AWR x BWR x 16 16 ( Subtract mAWB m mAWB 26 4x from each side.3 . Divide each side by ) 48 16 16 16 for x. 16 m AWR 16 for m⬔BWR. Substitute mBWR 4 4x 48 Substitute x for m⬔AWR and 4x 48 mAWR angle bisector Definition of 48 3x X A and mBWR 4x 48. Find mAWB. x Angle Addition BWR 32 Substitute 16 All rights reserved. All rights reserved. . 6 North 4Central Cedar 2 represent Oak. (x2, y2 ) represent 4 2 Oak (1, 2) ( ( 1 d% 2 2 4 2 d% 4 20 16 ))2 ( 2 ( Elm Use the Distance Formula. 2 ))2 Jackson (2, 4) Symphony City Plaza (0, 0) x 6 Substitute. and (x 2, y 2 ) be , y1 y2 ) 6 2 ) 1 Substitute 8 for x1 and 6 for x2. Simplify. 3 ) 3 Substitute 9 for y1 and 3 fory2. Simplify. (1, 3) . 2 The y-coordinate is 8 ( 2 9 ( Midpoint Formula . 2 2 6 2 x 1x y 1y a 1 2 2 , 1 2 2 b be coordinates of G. (1, 4) and the midpoint x2 and y2 d Use the Midpoint Formula. S 5 1 x2 d Multiply each side by 2 . S 10 (1, 5) . Solve for , the Find the x-coordinate of G. 1 1 x2 2 2 3 x2 d Simplify.S 6 4 y2 2 4 y2 x2 The coordinates of G are (3, 6) . Simplify within parentheses. 20 %www 4.472135955 Simplify. Use a calculator. To the nearest tenth, the subway ride from Oak to Symphony is 28 y 4 South "(x2 2 x1) 2 1 (y2 2 y1) 2 d % 1 x2 2 Use the Midpoint Formula. Let (x 1, y 1 ) be Symphony. d(8, 9) D(1, 4). Find the coordinates of the other endpoint G. (x1, y1 ) . Let 27 3 Finding an Endpoint The midpoint of DG is M(1, 5). One endpoint is . ride from Oak to Symphony? Round to the nearest tenth. Each unit represents 1 mile. (1, 2) The coordinates of the midpoint M are 1 Applying the DistanceFormula How far is the subway Symphony has coordinates x1 The x-coordinate is Examples. Oak has coordinates (1, 2) . Let Geometry Lesson 1-7 The midpoint has coordinates ) 2 100° L Local Standards: ____________________________________ (y2 y1 )2 , mJKL 50° (6, 3) . ( y1 y2 mNKL Use theMidpoint Formula. Let (x 1, y 1 ) be © Pearson Education, Inc., publishing as Pearson Prentice Hall. , 2 K coordinates of its midpoint M. © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. x1 x2 J 2 Finding the Midpoint AB hasendpoints (8, 9) and (6, 3). Find the Formula: The Midpoint Formula The coordinates of the midpoint M of AB with endpoints A(x 1, y 1 ) and B(x 2, y 2 ) are the following: ( ) 4. If mJKN 50°, then find mNKL and mJKL. KN bisects JKL. NAEP 2005 Strand: Measurement Topic: Measuring Physical AttributesFormula: The Distance Formula The distance d between two points A(x 1, y 1 ) and B(x 2, y 2 ) is M T Name_____________________________________ Class____________________________ Date ________________ Key Concepts. S Daily Notetaking Guide The Coordinate Plane (x2 x1 )2 Z NDaily Notetaking Guide Lesson 1-8 Y 3. Draw ST. Construct its perpendicular bisector. for m⬔AWR and for m⬔BWR. Name_____________________________________ Class____________________________ Date________________ d Measure ⬔XYP and ⬔PYZ to see that they are congruent. PPostulate Geometry Lesson 1-7 Lesson Objectives 1 Find the distance between two points in the coordinate plane 2 Find the coordinates of the midpoint of a segment in the coordinate plane F bisector of XYZ. © Pearson Education, Inc., publishing as Pearson Prentice Hall. XY AB at the midpoint of AB,so XY is the perpendicular bisector of AB. B b. Explain how you can use your protractor to check that YP is the angle B © Pearson Education, Inc., publishing as Pearson Prentice Hall. A B ) X * ) * All rights reserved. A All rights reserved. X Y * ) S 2. a. Construct F with mF 2mB. Step 2 With the compasssetting, put the compass same point on point B and draw another long arc. Label the intersect points where the two arcs as X and Y. * ) R Y B Geometry Lesson 1-8 All-In-One Answers Version A 4.5 miles. Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 1-8 Geometry 29 7 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Quick Check. Lesson 1-9 1. a. Using the map in Example 1, find the distance between Elm and Symphony. Perimeter, Circumference, and Area Lesson Objectives about 8.9 miles 1 2 NAEP 2005 Strand: Measurement Find perimeters of rectangles and squares, and circumferences of circles Find areas of rectangles, squares,and circles Topic: Measuring Physical Attributes Local Standards: ____________________________________ All rights reserved. b. Maple is located 6 miles west and 2 miles north of City Plaza. Find the distance between Cedar and Maple. about 3.2 miles All rights reserved. Key Concepts. Perimeterand Area s b h s r dO h b Square with side length s. Perimeter P Area A 4s s2 Rectangle with base b and height h. C Circle with radius r and diameter d. Perimeter P Circumference C Area A 2b 2h pd or C 2pr bh Area 2. Find the coordinates of the midpoint of XY with endpoints X(2, 5) and Y(6, 13). pr 2The area of a region is the sum of the areas of its non-overlapping parts. Examples. 1 Finding Circumference G has a radius of 6.5 cm. Find the circumference of G in terms of π. Then find the circumference to the nearest tenth. C2 p Formula for circumference of a circle ( 6.5 C 13 π C 13 ) 6.5 Substitutefor r. Exact answer 40.840704 13p Use a calculator. , or about 40.8 cm. Daily Notetaking Guide Daily Notetaking Guide 6.5 cm G r The circumference of G is Geometry Lesson 1-8 . Postulate 1-10 C 2π 30 equal If two figures are congruent, then their areas are 31 Geometry Lesson 1-9Name_____________________________________ Class____________________________ Date________________ Date ________________ Name_____________________________________ Class____________________________ Date ________________ 2 Finding Perimeter in the CoordinatePlane 4 Finding Area of an Irregular Shape Find the area of the figure C Draw a horizontal line to separate the figure into three rectangle non-overlapping figures: one and two 8 AR 4 2 6 8 10 Find the length of each side. Add the lengths to find the perimeter. 9 0 AB % 81 BC u 11 CD % ( %( u 144 12 92wwww 0 2 %www 12 w2 ) 225 w %wwww u u 2 u 2 )2 ( 0 12 )2 9 11 2 0 144 uu 2 u Use the Distance Formula. 15 2 Ruler Postulate BC CD DA Perimeter 15 2 15 Perimeter 34 The perimeter of quadrilateral ABCD is 2 34 units. 3 Finding Area of a Circle Find the area of B in terms of π. Aπ A 8 ( 5 )2 2525 Add the areas. Simplify. 125 125 2 ft . Quick Check. b. Find the circumference of a circle with a diameter of 18 m to the nearest tenth. 56.5m 2. Graph quadrilateral KLMN with vertices K(3, 3), L(1, 3), M(1, 4), and N(3, 1). Find the perimeter of KLMN. 20 units 4 N(3, 1) y M(1, 4) 2 x O 4 2 3. You aredesigning a rectangular banner for the front of a museum. The banner will be 4 ft wide and 7 yd high. How much material do you need in square yards? 2 4 2 K(3, 3) 4 L(1, 3) 2 91 3 yd Formula for the area of a circle 1.5 )2 2.25 π The area of B is 32 yd. B r2 ( 25 d Simplify. S 36p m Aπ A d Substitute. SUse the Distance Formula. 225 w %wwww AB 1.5 75 2 1. a. Find the circumference of a circle with a radius of 18 m in terms of π. )2 Perimeter In B, r A )( 5 ) The area of the figure is Ruler Postulate 12 15 75 s 15 9 2 % 81 DA )2 ( © Pearson Education, Inc., publishing as Pearson Prentice Hall. ( ©Pearson Education, Inc., publishing as Pearson Prentice Hall. AB % All rights reserved. 4 ( All rights reserved. D 2 x 12 5 ft 5 ft d Use the formulas. S AS bh 5 ft . Find each area. Then add the areas. 6 A O squares 10 ft Draw and label ABCD on a coordinate plane. 15 ft to the right. 5 ft B 5 ft y 10 10 ft 12Quadrilateral ABCD has vertices A(0, 0), B(9, 12), C(11, 12), and D(2, 0). Find the perimeter. Substitute 1.5 for r. 1.5 yd 4. Copy the figure in Example 4. Separate it in a different way. Find the area. 125 ft2 2.25p Geometry Lesson 1-9 Geometry yd 2. Daily Notetaking Guide All rights reserved. (6, 9)Postulate 1-9 Daily Notetaking Guide Geometry Lesson 1-9 33 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3. The midpoint of XY has coordinates (4, 6). X has coordinates (2, 3). Find the coordinates of Y. © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. (4, 4) Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________Name_____________________________________ Class____________________________ Date ________________ Lesson 2-1 2 Finding a Counterexample Find a counterexample to show that this Conditional Statements Lesson Objectives 1 Recognize conditional statements Write converses ofconditional statements 2 conditional is false: If x 2 0, then x 0. NAEP 2005 Strand: Geometry If p , pSq then its measure is 180. then q . If the measure of an angle is If q , qSp All rights reserved. If an angle is a straight angle, You read it All rights reserved. Because (1) 2 1, which is greater than 0, thehypothesis is Symbolic Form 180, then it is a straight angle. then p . Because 1 0, the conclusion is false . true . The counterexample shows that the conditional is . false 3 Writing the Converse of a Conditional Write the converse of the following conditional. If x 9, then x 3 12. The converse of a conditionalexchanges the hypothesis and the conclusion. A conditional is an if-then statement. Conditional The truth value of a statement is “true” or “false” according to whether the statement is true or false, respectively. The converse of the conditional “if p, then q” is the conditional “if q, then p.” Examples. 1Identifying the Hypothesis and the Conclusion Identify the hypothesis and conclusion: If two lines are parallel, then the lines are coplanar. In a conditional statement, the clause after if is the hypothesis and the clause after then is the conclusion. © Pearson Education, Inc., publishing as Pearson PrenticeHall. The conclusion is the part of an if-then statement (conditional) that follows then. © Pearson Education, Inc., publishing as Pearson Prentice Hall. The hypothesis is the part that follows if in an if-then statement. All rights reserved. and . This counterexample must be an example in which x 2 0(hypothesis true) and x 0 or x 0 (conclusion false). Because any negative number has a positive square, one possible counterexample is 1. Conditional Statements and Converses Statement Example Converse false the conclusion is Local Standards: ____________________________________Vocabulary and Key Concepts. Conditional true A counterexample is a case in which the hypothesis is Topic: Mathematical Reasoning Converse Hypothesis Conclusion Hypothesis Conclusion x9 x 3 12 x 3 12 x9 So the converse is: If x 3 12, then x 9. Quick Check. 1. Identify the hypothesis and theconclusion of this conditional statement: If y 3 5, then y 8. Hypothesis: y 3 5 Conclusion: y 8 2. Show that this conditional is false by finding a counterexample: If the name of a state includes the word “New,” then the state borders an ocean. Counterexample: Hypothesis: Two lines are parallel. New Mexicodoes not border an ocean. It is a counterexample, so the conditional is false. Conclusion: The lines are coplanar. 34 Geometry Lesson 2-1 Daily Notetaking Guide Daily Notetaking Guide 35 Name_____________________________________ Class____________________________Date________________ Name_____________________________________ Class____________________________ Date ________________ Example. Lesson 2-2 4 Finding the Truth Value of a Converse Write the converse of the Biconditionals and Definitions Lesson Objectives 1 Writebiconditionals 2 Recognize good definitions conditional. Then determine the truth value of each. If a 2 25, then a 5. NAEP 2005 Strand: Geometry Topics: Dimension and Shape; Mathematical Reasoning Local Standards: ____________________________________ Conditional: If a2 25, then a 5.Vocabulary and Key Concepts. The converse exchanges the and the hypothesis conclusion . Converse: If a 5, then a2 25. Biconditional Statements A biconditional combines p S q and q S p as p 4 q. . A counterexample is a 5: (5) 2 25, and 5 5. Because 52 25, the converse is true . All rights reserved.false All rights reserved. The conditional is Statement Biconditional Example Symbolic Form An angle is a straight angle if You read it p if and p4q only if q . and only if its measure is 180°. A biconditional statement is the combination of a conditional statement and its converse. Quick Check. Abiconditional contains the words “if and only if.” 3. Write the converse of the following conditional: If two lines are not parallel and do not intersect, then they are skew. Examples. If two lines are skew, then they are not parallel and do not intersect. 1 Writing a Biconditional Consider the true conditionalstatement. Write its 4. Write the converse of each conditional. Determine the truth value of the conditional and its converse. (Hint: One of these conditionals is not true.) a. If two lines do not intersect, then they are parallel. Converse: If two lines are parallel, then they do not intersect. The conditional is falseand the converse is true . b. If x 2, then u x u 2. Converse: © Pearson Education, Inc., publishing as Pearson Prentice Hall. converse. If the converse is also true, combine the statements as a biconditional. © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc.,publishing as Pearson Prentice Hall. Geometry Lesson 2-1 Conditional: If x 5, then x 15 20. To write the converse, exchange the hypothesis and conclusion. Converse: If x 15 20, then x 5. When you subtract 15 from each side to solve the equation, you get x 5. Because both the conditional and itsconverse are a biconditional using the phrase true , you can combine them in if and only if . Biconditional: x 5 if and only if x 15 20. 2 Separating a Biconditional into Parts Write the two statements that form this biconditional. Biconditional: Lines are skew if and only if they are noncoplanar. Write abiconditional as two conditionals that are converses of each other. Conditional: If lines are skew, then they are noncoplanar. If u x u 2, then x 2. Converse: If lines are noncoplanar, then they are skew. The conditional is 36 true and the converse is false Geometry Lesson 2-1 All-In-One Answers Version A .Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-2 Geometry 37 9 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________Name_____________________________________ Class____________________________ Date ________________ 2. Write two statements that form this biconditional about integers greater than 1: 3 Writing a Definition as a Biconditional Show that this definition of triangle is reversible. Then write itas a true biconditional. A number is prime if and only if it has only two distinct factors, 1 and itself. Definition: A triangle is a polygon with exactly three sides. Statement: true The original conditional is . If a number is prime, then it has only two distinct factors, 1 and itself. Conditional: If a polygon is atriangle, then it has exactly three sides. true The converse is also . Converse: If a polygon has exactly three sides, then it is a triangle. Statement: , they can be combined to If a number has only two distinct factors, 1 and itself, then it is prime. Biconditional: A polygon is a triangle if and only if it has exactlythree sides. 4 Identifying a Good Definition Is the following statement a good All rights reserved. true All rights reserved. Because both statements are form a biconditional. definition? Explain. An apple is a fruit that contains seeds. 3. Show that this definition of right angle is reversible. Then write it as atrue biconditional. Definition: A right angle is an angle whose measure is 90. Conditional: The statement is true as a description of an apple. If an angle is a right angle, then its measure is 90°. Exchange “An apple” and “a fruit that contains seeds.” The converse reads: A fruit that contains seeds is anapple. . false is not The original statement statement is not , so the converse of the a good definition because the reversible. Quick Check. 1. Consider the true conditional statement. Write its converse. If the converse is also true, combine the statements as a biconditional. Conditional: If three points arecollinear, then they lie on the same line. Converse: If three points lie on the same line, then they are collinear. true The converse is . Converse: If an angle has measure 90°, then it is a right angle. The two statements are true All rights reserved. counterexamples statement is © Pearson Education, Inc.,publishing as Pearson Prentice Hall. and peaches. These are © Pearson Education, Inc., publishing as Pearson Prentice Hall. There are many fruits that contain seeds but are not apples, such as lemons . Biconditional: An angle is a right angle if and only if its measure is 90°. 4. Is the following statementa good definition? Explain. A square is a figure with four right angles. It is not a good definition because a rectangle has four right angles and is not necessarily a square. Biconditional: Three points are collinear if and only if they lie on the same line. 38 Geometry Lesson 2-2 Daily Notetaking Guide DailyNotetaking Guide Geometry Lesson 2-2 39 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 2-3 3 Usingthe Law of Syllogism Use the Law of Syllogism to draw a Deductive Reasoning conclusion from the following true statements: If a quadrilateral is a square, then it contains four right angles. If a quadrilateral contains four right angles, then it is a rectangle. NAEP 2005 Strand: Geometry Topic: MathematicalReasoning Local Standards: ____________________________________ The conclusion of the first conditional is the hypothesis of the second Law of Syllogism conditional. This means that you can apply the Vocabulary and Key Concepts. The Law of Syllogism: If then is true. In symbolic form: If p S q isa true statement and p is true, then q is true. Law of Syllogism All rights reserved. conclusion All rights reserved. Law of Detachment If a conditional is true and its hypothesis is true, then its © Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson Objectives 1 Use the Law of Detachment 2Use the Law of Syllogism pSr pSq and qSr . are true statements, is a true statement. So you can conclude: If a quadrilateral is a square, then it is a rectangle. 4 Drawing Conclusions Use the Laws of Detachment and Syllogism to draw a possible conclusion. If the circus is in town, then there are tents atthe fairground. If there are tents at the fairground, then Paul is working as a night watchman. The circus is in town. If p S q and q S r are true statements, then p S r is a true statement. Deductive reasoning is a process of reasoning logically from given facts to a conclusion. Because the conclusion of thefirst statement is the of the second statement, you can apply the hypothesis to write Law of Syllogism a new conditional: 1 Using the Law of Detachment A gardener knows that if it rains, the garden will be watered. It is raining. What conclusion can he make? The first sentence contains a conditionalstatement. The hypothesis is it rains. Because the hypothesis is true, the gardener can conclude that the garden will be watered. 2 Using the Law of Detachment For the given statements, what can you conclude? Given: If A is acute, then mA 90°. A is acute. A conditional and its hypothesis are both givenas true. By the Law of Detachment conclusion of the conditional, mA 90°, is , you can conclude that the true . © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Examples. is working as a night watchman. If the circus is in town,then Paul ________________________________________ The third statement means that the hypothesis of the new conditional is true. You can use the Law of Detachment to form the conclusion: Paul is working as a night watchman. Quick Check. 1. Suppose that a mechanic begins work on a car andfinds that the car will not start. Can the mechanic conclude that the car has a dead battery? Explain. No, there could be other things wrong with the car, such as a faulty starter. 2. If a baseball player is a pitcher, then that player should not pitch a complete game two days in a row. Vladimir Nuñez is apitcher. On Monday, he pitches a complete game. What can you conclude? Answers may vary. Sample: Vladimir Nuñez should not pitch a complete game on Tuesday. 40 10 Geometry Lesson 2-3 Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-3 41 All-In-One AnswersVersion A Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Lesson 2-4 3. If possible, state a conclusion using the Law of Syllogism. If it is not possible to use this law, explain why. a. If a number ends in 0, then it is divisible by 10. If a number is divisible by 10, then it is divisible by 5. Reasoning in Algebra Lesson Objective 1 NAEP 2005

Strand: Algebra and Geometry Connect reasoning in algebra and geometry Topics: Algebraic Representations; Mathematical Reasoning If a number ends in 0, then it is divisible by 5. Local Standards: ____________________________________ All rights reserved. b. If a number ends in 6, then it isdivisible by 2. If a number ends in 4, then it is divisible by 2. Not possible; the conclusion of one statement is not the hypothesis of the other statement. All rights reserved. Key Concepts. Properties of Equality Addition Property If a b, then a c b c . Subtraction Property If a b, then a c b c . MultiplicationProperty If a b, then a c b c . Division Property © Pearson Education, Inc., publishing as Pearson Prentice Hall. The Volga River is in Europe. If a river is less than 2300 miles long, it is not one of the world’s ten longest rivers. If a river is in Europe, then it is less than 2300 miles long. Conclusion: All rightsreserved. The Volga River is less than 2300 miles long. Conclusion: The Volga River is not one of the world’s ten longest rivers. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4. Use the Law of Detachment and the Law of Syllogism to draw conclusions. If a b and c 0, then a c b . cReflexive Property a a . Symmetric Property If a b, then b a . Transitive Property If a b and b c, then a c . Substitution Property If a b, then b can replace a in any expression. Distributive Property a(b c) ab ac . Properties of Congruence Reflexive Property AB AB A ⬔A Symmetric Property If AB CD, thenCD AB . If A B, then B ⬔A . Transitive Property If AB CD and CD EF , then AB EF . If A B and B C, then A ⬔C . 42 Geometry Lesson 2-3 Daily Notetaking Guide Daily Notetaking Guide Examples. Example. 1 Justifying Steps in Solving an Equation Justify each step used to solve 3 Using Properties ofEquality and Congruence Name the property that 5x 12 32 x for x. justifies each statement. a. If x y and y 4 3x, then x 4 3x. Given: 5x 12 32 x 5x 44 x Addition Property of Equality 4x 44 Subtraction Property of Equality x 11 The conclusion of the conditional statement is the same as the equation y 4 3x(given) after x has been substituted for y (given). The property used is the Substitution Property of Equality. Division Property of Equality AB BC AC 19 4 2x B C Segment Addition Postulate 21 x 21 (4 2x) A All rights reserved. 15 x points A, B, and C are collinear with point B between points A and C. Solvefor x if AC 21, AB 15 x, and BC 4 2x. Justify each step. All rights reserved. b. If x 4 3x, then 4 2x. 2 Justifying Steps in Solving an Equation Suppose that (15 x) 43 Name_____________________________________ Class____________________________ Date ________________ c. If P Q, Q R, and RS, then P S. for Transitive Property of Congruence the first two parts of the hypothesis: Simplify. x 2 The conclusion of the conditional statement shows the result after x is from each side of the equation in the hypothesis. subtracted The property used is the Subtraction Property of Equality. Use theSubstitution Property of Equality If P Q and Q R, then Subtraction Property of Equality Use the ⬔P 艑 ⬔R . Transitive Property of Congruence for P R and the third part of the hypothesis: Quick Check. 1. Fill in each missing reason. ) M Given: LM bisects KLN. (2x40) 4x ) K LM bisects KLN mMLN mKLMDefinition of Angle N Bisector 4x 2x 40 Substitution Property of Equality 2x 40 Subtraction Property of Equality x 20 L Given Division Property of Equality 15 x 2. Recall that the length of line AC is 21. A 4 2x B C © Pearson Education, Inc., publishing as Pearson Prentice Hall. If P R and R S, then ©Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Name_____________________________________ Class____________________________ Date________________ Geometry Lesson 2-4 ⬔P 艑 ⬔S . The property used isthe Transitive Property of Congruence. Quick Check. 3. Name the property of equality or congruence illustrated. a. XY XY Reflexive Property of Congruence b. If mA 45 and 45 mB, then mA mB Transitive or Substitution Property of Equality Find the lengths of AB and BC by substituting x 2 into theexpressions in the diagram. Check that AB BC 21. AB 44 13 BC 8 AB BC 13 8 Geometry Lesson 2-4 All-In-One Answers Version A 21 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 2-4 Geometry 45 11 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date________________ Lesson 2-5 Proving Angles Congruent Lesson Objectives complementary angles supplementary angles NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures 2Local Standards: ____________________________________ ⬔1 and ⬔2 are complementary angles. Vocabulary and Key Concepts. Theorem 2-1: Vertical Angles Theorem 3 4 2 Theorem 2-2: Congruent Supplements Theorem If two angles are supplements of the same angle (or of congruent angles),congruent All rights reserved. . 1 2 and 3 4 then the two angles are congruent Theorem 2-4 right All A theorem is a conjecture that is proven. A paragraph proof is a convincing argument that uses deductive reasoning in which statements and reasons are connected in sentences. Examples. adjacentangles 1 4 3 2 1 ⬔1 and ⬔2 are vertical angles, 2 4 ⬔1 and ⬔2 are adjacent angles, as are ⬔3 and ⬔4 . ⬔4 . © Pearson Education, Inc., publishing as Pearson Prentice Hall. vertical angles angle. right © Pearson Education, Inc., publishing as Pearson Prentice Hall. Theorem 2-5 If two angles arecongruent and supplementary, then each is a as are ⬔3 and Two angles are supplementary angles if the sum of their measures is 180°. . angles are congruent. 3 Two angles are complementary angles if the sum of their measures is 90°. . Theorem 2-3: Congruent Complements Theorem If two angles arecomplements of the same angle (or of congruent angles), then the two angles are ⬔3 and ⬔4 are supplementary angles. 1 All rights reserved. congruent Vertical angles are 1 Using the Vertical Angles Theorem Find the value of x. The angles with labeled measures are vertical angles. Apply the VerticalAngles Theorem to find x. 4x 101 2x 3 Vertical Angles Theorem 4x 2x 104 2x 104 Subtraction Property of Equality x 52 Division Property of Equality 2 Proving Theorem 2-2 Write a paragraph proof of Theorem 2-2 using the diagram at the right. Start with the given: 1 and 2 are supplementary, 3 and 2 aresupplementary angles supplementary. By the definition of Adjacent angles are two coplanar angles m1 m2 180 and m3 m2 180. By substitution, sides form two pairs of opposite rays. that have a common side and a common m1 m2 vertex but no common interior points. m⬔2 Geometry Lesson 2-5 m⬔3m⬔3 Name_____________________________________ Class____________________________ Date ________________ 2 3 , , subtract m⬔2 from , or ⬔ 1 ⬔ 3 . Daily Notetaking Guide Daily Notetaking Guide 1 . Using the Subtraction Property of Equality each side. You get m1 (2x 3) (4x 101)Addition Property of Equality Vertical angles are two angles whose 46 4 3 1 All rights reserved. Prove and apply theorems about angles 47 Geometry Lesson 2-5 Name_____________________________________ Class____________________________ Date________________ Lesson 3-1 QuickCheck. Properties of Parallel Lines Lesson Objectives 1 Identify angles formed by two lines and a transversal 2 Prove and use properties of parallel lines 1. Refer to the diagram for Example 1. a. Find the measures of the labeled pair of vertical angles. 107° b. Find the measures of the other pair of verticalangles. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes Local Standards: ____________________________________ Vocabulary and Key Concepts. c. Check to see that adjacent angles are supplementary. 107° 73° 180° All rights reserved. All rights reserved. 73° Postulate 3-1:Corresponding Angles Postulate If a transversal intersects two parallel lines, then corresponding congruent angles are . ⬔1 12 Geometry Lesson 2-5 Geometry Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as PearsonPrentice Hall. No, the size of the angles does not affect the proof or the truth of the theorem. 48 1 2 / m ⬔1 ⬔2 Theorem 3-1: Alternate Interior Angles Theorem If a transversal intersects two parallel lines, then alternate interior congruent angles are . 2. Recall the proof of Theorem 2-2. Does the size of theangles in the diagram affect the proof? Would the proof change if 1 and 3 were acute rather than obtuse? Explain. t ⬔3 a b t 4 3 2 1 5 6 Theorem 3-2: Same-Side Interior Angles Theorem If a transversal intersects two parallel lines, then same-side interior angles are supplementary . m1 m2 180 Theorem3-3: Alternate Exterior Angles Theorem If a transversal intersects two parallel lines, then alternate exterior angles are congruent . ⬔4 ⬔5 Theorem 3-4: Same-Side Exterior Angles Theorem If a transversal intersects two parallel lines, then same-side exterior angles are supplementary . m4 m6 180 DailyNotetaking Guide Geometry Lesson 3-1 49 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 Name_____________________________________ Class____________________________ Date ________________ Geometry: All-In-One Answers Version A(continued) Name_____________________________________ Class____________________________ Date ________________ A transversal is a line that intersects two coplanar lines at two m Alternate interior angles are nonadjacent interior angles that lie In the diagram at right, ᐉm and pq. Find m1and m2. 1 and the 42 angle are corresponding angles Because ᐉm, m1 by the on opposite sides of the transversal. ⬔1 the steps 42 3 Using Algebra to Find Angle Measures 65 Alternate Interior Angles Theorem c 40 Alternate Interior Angles Theorem abc 65 b b 3 1 The angle vertical to 1 is between therunway segments. 2 is between the runway segments and on the opposite side of the alternate interior angles are not adjacent and lie between the lines on opposite sides of the transversal, 2 and the angle vertical to 1 are alternate interior angles . © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. 2 180 / 40 65 Angle Addition 180 40 1 Applying Properties of Parallel Lines In the diagram of Lafayette Regional Airport, the black segments are runways and the gray areas are taxiways and terminalbuildings. . a b c In the diagram, ᐉm. Find the values of a, b, and c. a Compare 2 and the angle vertical to 1. Classify the angles as alternate interior angles, same-side interior angles, or corresponding angles. / m m⬔2 180 . 42 The first Examples. transversal runway. Because Angle Addition Postulate form1, the equation becomes shows the reason for each step. the steps to prove a theorem. and the second column All rights reserved. ⬔1 and ⬔7 corresponding angles. lines. 5 42 from each side to find m2 138 . 42 Subtract All rights reserved. angles that lie on the same side of the transversal and incorresponding positions relative to the coplanar column shows 4 . straight angle, m1 m2 180 by the If you substitute and ⬔4 are same-side interior angles. q . 8 6 7 2 1 3 Because 1 and 2 are adjacent angles that form a ⬔1 and ⬔2 are alternate interior angles. of the transversal. A two-column proof is adisplay that shows 42 Corresponding Angles Postulate Same-side interior angles are interior angles that lie on the same side Corresponding angles are p 2 Finding Measures of Angles t 56 13 42 78 / distinct points. Name_____________________________________Class____________________________ Date ________________ m Postulate Substitution Property of Equality 75 Subtraction Property of Equality Quick Check. 1. Use the diagram in Example 1. Classify 2 and 3 as alternate interior angles, same-side interior angles, or corresponding angles. same-sideinterior angles 2. Using the diagram in Example 2 find the measure of each angle. Justify each answer. a. 3 42; Vertical angles are congruent b. 4 138; Corresponding angles are congruent c. 5 138; Same-side interior angles are supplementary d. 6 42; Corresponding angles are congruent e. 7 138;Alternate interior angles are congruent f. 8 138; Vertical angles are congruent 3. Find the values of x and y. Then find the measures of the four angles in the trapezoid. 50 Geometry Lesson 3-1 (y 50)° Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________Class____________________________ Date________________ Lesson Objectives 1 Use a transversal in proving lines parallel Geometry Lesson 3-1 51 Name_____________________________________ Class____________________________ Date ________________ Proving Lines Parallel A flowproof uses arrows to show the logical connections between the statements. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes Reasons are written below the statements. Local Standards: ____________________________________ Vocabulary and Key Concepts. Examples.Theorem 3-5: Converse of the Alternate Interior Angles Theorem If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel. 2 ᐉ m . All rights reserved. m then the two lines are parallel. All rights reserved. If two lines and a transversal form correspondingangles that are congruent, 1 Using Postulate 3-2 1 / Postulate 3-2: Converse of the Corresponding Angles Postulate 5 6 1 4 2 m 3 If 1 2, / supplementary, supplementary, then the two lines are parallel. then ᐉ m . Theorem 3-7: Converse of the Alternate Exterior Angles Theorem If 3 5, If two lines and atransversal form alternate exterior angles that are then ᐉ m . congruent, then the two lines are parallel. Theorem 3-8: Converse of the Same-Side Exterior Angles Theorem If 3 and 6 are If two lines and a transversal form same-side exterior angles that are supplementary, supplementary, then the two linesare parallel. then ᐉ m . 52 Geometry Lesson 3-2 All-In-One Answers Version A Daily Notetaking Guide and ⬔2 D1 are supplementary. Because 4K 2 congruent supplements of the same angle are ⬔4 . Because Converse of the Corresponding Angles ) DK ) ⬔3 by the Postulate. 2 Using Algebra ©Pearson Education, Inc., publishing as Pearson Prentice Hall. If 2 and 4 are If two lines and a transversal form same-side interior angles that are shows that ⬔4 3 C E It is given that 3 and 2 are supplementary. The diagram and ⬔4 are congruent corresponding angles, EC then ᐉ m . Theorem 3-6:Converse of the Same-Side Interior Angles Theorem Use the diagram at the right. Which lines, if any, must be parallel if 3 and 2 are supplementary? Justify your answer with a theorem or postulate. (Congruent Supplements Theorem), ⬔3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. ©Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson 3-2 y° 2x° x 45, y 115; 90, 90, 115, 65 Find the value of x for which ᐉm. The labeled angles are alternate interior angles . If ᐉm, the alternate interior angles are , equal and their measures are equation 5x 5x 66 14 congruent / (5x 66) .Write and solve the m 3x. 66 14 3x 5x 80 3x 2x 80 Subtract x 40 Divide each side by 2 . Daily Notetaking Guide (14 3x) Add 66 to each side. 3x from each side. Geometry Lesson 3-2 Geometry 53 13 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date________________ Quick Check. Lesson 3-3 1. Use the diagram from Example 1.Which lines, if any, must be parallel if 3 4? Explain. Parallel and Perpendicular Lines Lesson Objectives 1 u u EC 储 DK ; Converse of Corresponding Angles Postulate NAEP 2005 Strand: Geometry Relate parallel and perpendicular lines Topic: Relationships Among Geometric Figures Local Standards:____________________________________ a x 18; 7(18) 8 118°, and 62° 118° 180° Key Concepts. (7x 8) Theorem 3-9 62 All rights reserved. b All rights reserved. 2. Find the value of x for which a b. Explain how you can check your answer. parallel to the same line If two lines are they are parallel a b ,then c to each other. a b . perpendicular to the same line In a plane, if two lines are then they are t m Theorem 3-10 parallel , n to each other. Geometry Lesson 3-2 Name_____________________________________ Class____________________________ Date ________________ side are parallel.And if an outer inner inner and outer Theorem 3-9 outer outer opposite side. But that outer edge is parallel to the inner All rights reserved. Examples. 1 Real-World Connection A picture frame is assembled as shown. Given the book’s explanation for why the outer edges on opposite sides of the frame areparallel, why must the inner edges on opposite sides be parallel, too? corners cut to form 45 angles 55 Geometry Lesson 3-3 Parallel Lines and the Triangle Angle-Sum Theorem Lesson Objectives 1 Classify triangles and find the measures of their angles 2 Use exterior angles of triangles edge on theopposite side, then by outer ᐉ m Lesson 3-4 edge on one side is parallel to the each inner edge is parallel to the n m Name_____________________________________ Class____________________________ Date________________ edges of each edge on the same side, and at the same time theedge is parallel to the ⊥ Daily Notetaking Guide Daily Notetaking Guide The sides are constructed so that the n edge on the NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures Local Standards: ____________________________________ edge on the same side, so again byTheorem 3-9 , inner edges on opposite sides are Vocabulary and Key Concepts. 1 k Write a paragraph proof. ᐉ Given: In a plane, k l and k m. Also, m1 90°. m Prove: The transversal is perpendicular to line m. All rights reserved. 2 Using Theorem 3-11 All rights reserved. parallel. Since m⬔1 90° , thetransversal is perpendicular to line k . Theorem 3-12: Triangle Angle-Sum Theorem The sum of the measures of the angles of a triangle is 180 C . mA mB mC 180 B A Theorem 3-13: Triangle Exterior Angle Theorem The measure of each exterior angle of a triangle equals the sum of the 2 1 measures ofits two remote interior angles. Since k 储 m , by Theorem 3-11 the transversal is also 3 perpendicular to line m. Yes; 30° 60° 90° 60 60 60 60 30 30 2. From what is given in Example 2, can you also conclude that the transversal is perpendicular to line l? Yes, by Theorem 3-11 and the fact that k 储 l ©Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. Can you assemble the framing at the right into a frame with opposite sides parallel? Explain. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Quick Check. acute obtuse right An A An triangle has triangle has triangle has oneobtuse angle. three acute angles. one right angle. __________________ __________________ __________________ __________________ __________________ __________________ An equilateral triangle has three congruent sides. ________________________ ________________________isosceles An triangle has at least two congruent sides. __________________________ __________________________ An equiangular triangle has three congruent angles. ______________________ ______________________ scalene A triangle has no congruent sides. __________________________________________ An exterior angle of a polygon is an angle formed by a side 1 and an extension of an adjacent side. Remote interior angles are the two nonadjacent interior angles corresponding to each exterior angle of a triangle. 56 14 Geometry Lesson 3-3 Geometry Daily Notetaking GuideDaily Notetaking Guide Exterior angle 2 3 Remote interior angles Geometry Lesson 3-4 57 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 54 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as PearsonPrentice Hall. m n . Theorem 3-11 In a plane, if a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other. Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________Date________________ Name_____________________________________ Class____________________________ Date ________________ Examples. Quick Check. 1 Applying the Triangle Angle-Sum Theorem In triangle ABC, ACB is a right angle, and CDAB. Find the values of a and c. 90 70 cAngle Addition 20 Subtract 70 For 䉭CAB, (70 20) 70 b 180. Then 160 b 180 and b 20. For 䉭CDB, 70 90 b 180. Then 160 b 180 and b 20. b a A a right angle Definition of B D Postulate from each side. 2. a. Find m3. a mADC c mADC a 90 a 20 110 a Triangle Angle-Sum 180 90 Substitute 180 Simplify.70 90 20 for m⬔ADC and 110 Subtract Theorem perpendicular lines Definition of 180 All rights reserved. To find a, use ADC. for c. All rights reserved. c 90 70 c Find c first, using the fact that ACB is a right angle. mACB 1. Critical Thinking Describe how you could use either CAB or CDB to find the valueof b in Example 1. C 45 90 45 b. Critical Thinking Is it true that if two acute angles of a triangle are complementary, then the triangle must be a right triangle? Explain. 3 True, because the measures of the two complementary angles add to 90, leaving 90 for the third angle. from each side. 2 Applying theTriangle Exterior Angle Theorem Explain what happens All rights reserved. x (180180 ) (180180 x ) x The exterior angle and the angle formed by the back of the chair and the adjacent angles armrest are straight angle decreases measure the armrest 58 , which together form a increases . As one measureincreases , the other . The angle formed by the back of the chair and as you make a lounge chair recline more. Geometry Lesson 3-4 © Pearson Education, Inc., publishing as Pearson Prentice Hall. x Arm © Pearson Education, Inc., publishing as Pearson Prentice Hall. Back to the angle formed by theback of the chair and the armrest as you make a lounge chair recline more. Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date ________________ The Polygon Angle-Sum Theorems Lesson Objectives 1 Classify polygons 2 Find the sums of the measures of the interior and exterior angles of polygons An equilateral polygon has all sides congruent. An equiangular polygon has allangles congruent. NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures A regular polygon is both equilateral and equiangular. Local Standards: ____________________________________ Examples. 1 Classifying Polygons Classify the polygon at the right by its sides. Identify itVocabulary and Key Concepts. as convex or concave. Starting with any side, count the number of sides clockwise around the figure. Because the polygon has 12 sides, it is a dodecagon. Think of the . Theorem 3-15: Polygon Exterior Angle-Sum Theorem The sum of the measures of the exterior angles ofa polygon, 360 one at each vertex, is 3 All rights reserved. (n 2)180 The sum of the measures of the angles of an n-gon is All rights reserved. Theorem 3-14: Polygon Angle-Sum Theorem polygon as a star. Draw a diagonal connecting two adjacent points of the star. That diagonal lies concave 4 360 1 . 2Finding a Polygon Angle Sum Find the sum of the measures of the angles 5 of a decagon. A polygon is a closed plane figure with at least three sides that are segments. The sides Sum (n 2)(180) intersect only at their endpoints, and no two adjacent sides are collinear. B B A A C E C E D not a D D Not apolygon A polygon C closed ; figure Not a polygon A concave polygon has at least one diagonal with points outside of the polygon. ; two sides intersect between endpoints. A convex polygon does not have diagonal points outside of the polygon. E A D R Y T Diagonal P M K W Q G S A convex polygon Aconcave polygon © Pearson Education, Inc., publishing as Pearson Prentice Hall. B sides, so n 10 A decagon has A the polygon, so the dodecagon is outside . 2 . For the pentagon, m1 m2 m3 m4 m5 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishingas Pearson Prentice Hall. Lesson 3-5 59 Geometry Lesson 3-4 ( 10 10 . Polygon Angle-Sum ) 2 (180) Substitute 8 ? 180 Subtract. Simplify. 1440 10 3 Using the Polygon Angle-Sum Theorem Find mX in quadrilateral Geometry Lesson 3-5 All-In-One Answers Version A Daily Notetaking Guide Y Z XYZW.The figure has 4 sides, so n 4 . 100° X mX mY mZ mW mX mY (4 2)(180) Polygon Angle-Sum 100 360 Substitute. mX mY 190 360 Simplify. mX mY 170 Subtract 170 Substitute 170 Simplify. 90 mX m⬔X 2m⬔X mX 60 Theorem for n. Daily Notetaking Guide 85 190 W Theorem from each side. m⬔X form⬔Y. Divide each side by 2 . Geometry Lesson 3-5 Geometry 61 15 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________Class____________________________ Date ________________ 4 Applying Theorem 3-15 A regular hexagon is inscribed in a rectangle. Lesson 3-6 Explain how you know that all the angles labeled 1 have equal measures. 2 2 1 Lines in the Coordinate Plane 1 Lesson Objectives congruent Thehexagon is regular, so all its angles are supplement exterior angle is the 1 2 1 , all the angles marked 1 congruent Graph lines given their equations Write equations of lines Topics: Patterns, Relations, and Functions; Algebraic Representations supplements are adjacent angles that form a straight angle.Because of congruent angles are . An of a polygon’s angle because they NAEP 2005 Strand: Algebra Local Standards: ____________________________________ 1 2 2 have equal measures. Vocabulary. Slope-intercept form All rights reserved. Quick Check. 1. Classify each polygon by its sides.Identify each as convex or concave. a. b. All rights reserved. The slope-intercept form of a linear equation is y 2x 3 y mx b. 1 Ax By C. Standard The point-slope form for a nonvertical line is form y y 1 m(x x 1 ). hexagon; convex O 4 2x y 1 4 3 octagon; concave x y 2 2(x 1) Point-slope form Examples. 2.a. Find the sum of the measures of the angles of a 13-gon. 1 Graphing Lines Using Intercepts Use the x-intercept and y-intercept to You can solve the equation (n 2)180 720. 3. Pentagon ABCDE has 5 congruent angles. Find the measure of each angle. 108 4. a. In the figure from Example 4, find m1 byusing the Polygon Exterior Angle-Sum Theorem. graph 5x 6y 30. To find the x-intercept, substitute 0 for y and solve for x. To find the y-intercept, substitute 0 for x and solve for y. 5x 6y 30 5x 6y 30 ( ) 30 ( ) 6y 30 5x 6 0 5 0 5x 0 30 5x 0 6y 30 6y 30 x 6 30 y 5 The y-intercept is 5 . The x-intercept is 6 . Apoint on the line is ( 6 , 0). ( A point on the line is 0, 5 Plot (6, 0) and (0, 5). Draw the line containing the two points. (6, 0) 6 Geometry Lesson 3-5 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ 3 y 4 2x 63 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3 Geometry Lesson 3-6 point-slope form of the line that contains the points G(4, 9) and H(1, 1). 6x 3y 12 3 (0, 5) 4 Writing an Equation of a Line Given Two Points Write an equation in Step 1 Transform theequation to slope-intercept form. 12 8x Example. 6x 3y 12 to slope-intercept form. Then graph the resulting equation. 6x 6 Name_____________________________________ Class____________________________ Date ________________ 2 Transforming to Slope-Intercept Form Transform theequation 3y 4 4 30; no, it is not formed by extending one side of the polygon. 12 2 2 b. Find m2. Is 2 an exterior angle? Explain. 6x ). y 2 2 O 60 3y All rights reserved. b. Critical Thinking The sum of the measures of the angles of a given polygon is 720. How can you use Sum (n 2)180 to find the number ofsides in the polygon? © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1980 62 y The standard form of a linear equation is Step 1 Find the slope. Add 6x to each side. m Divide each side by 3 . Simplify. 4 y (1, 6) (0, 4) Allrights reserved. 6 All rights reserved. Step 2 Use the y-intercept and the slope to plot two points and draw the line containing them. y1 x2 x1 1 m The y-intercept is 4 and the slope is 2 . y2 1 ( Use the formula for slope. 9 ) Substitute 4 (4, 9) for (x1, y1) and (1, 1) for (x2, y2). 10 Simplify. 5 2 2 Step 2 Selectone of the points. Write the equation in point-slope form. 2 O 2 4x y 2 y ( y1 9 ( m x ) 2 y 9 2 3 Using Point-Slope Form Write an equation in point-slope form x1 (x (x 4 4 ) ) ) Point-slope form. Substitute 2 for m and (4, 9) for (x1, y1). Simplify. ( y1 ( m x ) 6 8 y 6 8 x1 (x (x 3 3 ) ) ) Use point-slope form.Substitute 8 for m and (3, 6) for (x1, y1). Simplify. Quick Check. 1. Graph each equation. a. y 12x 2 b. 2x 4y 8 y O 1 x 16 1 2 4 c. 5x y 3 x 2 O 3 64 O y y 2 x 2 Geometry Lesson 3-6 Geometry Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. y y © PearsonEducation, Inc., publishing as Pearson Prentice Hall. of the line with slope 8 that contains P(3, 6). Quick Check. 2. Write an equation of the line with slope 1 that contains the point P(2, 4). y 4 1(x 2) 3. Write an equation of the line that contains the points P(5, 0) and Q(7, 3). 3 y 0 3 2(x 5) or y 3 2 (x 7) DailyNotetaking Guide Geometry Lesson 3-6 65 All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date ________________ Lesson 3-7 2 Writing Equations of Parallel Lines Write an equation in point-slope Slopes of Parallel and Perpendicular Lines Lesson Objectives 1 Relate slope and parallel lines Relate slope and perpendicular lines 2 form for the lineparallel to 6x 3y 9 that contains (5, 8). NAEP 2005 Strand: Measurement Step 1 To find the slope of the line, rewrite the equation in slope-intercept form. Topic: Measuring Physical Attributes 6x 3y 9 Local Standards: ____________________________________ 3 y 6x 9 y Key Concepts. Any two verticallines are parallel. Slopes of Perpendicular Lines is 1. slopes Examples. x 5y 4 x 4 5y 4 x 5 Subtract 4 from each side. y 5 5 . Divide each side by 1 y 4 x 5 5 The line x 5y 4 has slope 1 5 . The line y 5x 4 has slope 5 . are not parallel because their slopes are not equal © Pearson Education, Inc., publishingas Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. slope-intercept form. Write the equation x 5y 4 in slope-intercept form. The lines x1 ) ( ) 2 ( x ( 5 )) 2 (x 5 ) m x 8 y 8 Substitute 2 for m and (5, 8) for (x1, y1). Simplify. b. y 12x 5 and 2x 4y 20 No;the lines have the same slope and y-intercept, so they are the same line. 2. Write an equation for the line parallel to y x 4 that contains (2, 5). y 5 (x 2) . Geometry Lesson 3-7 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________Class____________________________ Date ________________ Geometry Lesson 3-7 67 Name_____________________________________ Class____________________________ Date________________ Lesson 3-8 Example. Lesson Objectives 1 Construct parallel lines 2 Constructperpendicular lines 3 Writing Equations for Perpendicular Lines Write an equation for a line perpendicular to 5x 2y 1 that contains (10, 0). Step 1 To find the slope of the given line, rewrite the equation in slope-intercept form. Constructing Parallel and Perpendicular Lines NAEP 2005 Strand: GeometryTopic: Relationships Among Geometric Figures Local Standards: ____________________________________ 5x 2y 1 2y 5 x 1 5x Example. from each side. 1 Constructing ᐉm Examine the diagram at right. Explain 2 5 2 The line 5x 2y 1 has slope . Step 2 Find the slope of a line perpendicular to 5x 2y 1.Let m be the slope of the perpendicular line. 5 2 m 1 m 1 The product of the slopes of perpendicular lines is ? ( ) 2 5 m y 0 y 2 5 2 5 2 Multiply each side by . 5 Simplify. 5 ( x 10 (x 10 y1 m(x x1 ), to write an 2 ) Substitute ) Simplify. for m and 5 (10, 0) how to construct 1 congruent to H. Construct the angle.N 1 Use the method learned for constructing congruent angles. m Step 1 With the compass point on point H, draw an arc that intersects the sides of ⬔H . / H J Step 2 With the same compass setting, put the compass point on point N . Draw an arc. 1 . 2 Step 3 Use point-slope form, y equation for the newline. All rights reserved. Divide each side by 2 . for (x1, y1). Quick Check. 3. Write an equation for the line perpendicular to 5y x 10 that contains (15, 4). © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2 Subtract 1 x All rights reserved. 5 y © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. y1 ( y-intercepts are different. x 5y 4 parallel? Explain. 66 y y Yes; Each line has slope 1 2 and the 1 3 . . a. y 12x 5 and 2x 4y 9 1 Determining Whether Lines are Parallel Are the lines y 5x 4 and 2 1. Are the lines parallel?Explain. Any horizontal line and vertical line are perpendicular. The equation y 5x 4 is in from each side. Quick Check. of two lines have a product of 1, the lines are perpendicular. slopes If the All rights reserved. of two distinct nonvertical lines are equal, the lines are parallel. If two nonvertical lines areperpendicular, the product of their 6x Subtract Divide each side by Step 2 Use point-slope form to write an equation for the new line. are equal. slopes All rights reserved. If two nonvertical lines are parallel, their slopes 3 The line 6x 3y 9 has slope Slopes of Parallel Lines If the 2x Step 3 Put the compasspoint below point N where the arc * ) intersects HN . Open the compass to the length where the arc intersects line ᐉ . Keeping the same compass setting, put the compass point above point N where the * ) arc intersects HN . Draw an arc to locate a point. Step 4 Use a straightedge to draw line m throughthe point you located and point N. Quick Check. 1. Use Example 1. Explain why lines ᐉ and m must be parallel. If corresponding angles are congruent, the lines are parallel by the Converse of Corresponding Angles Postulate. y 4 5(x 15) 68 Geometry Lesson 3-7 All-In-One Answers Version A DailyNotetaking Guide Daily Notetaking Guide Geometry Lesson 3-8 Geometry 69 17 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________Name_____________________________________ Class____________________________ Date ________________ Example. Example. 2 Constructing a Special Quadrilateral Construct a quadrilateral with both the 3 Perpendicular From a Point to a Line * Examine ) R construction. At what specialpoint does RG meet line ᐉ? pairs of sides parallel. Step 1 Draw point A and two rays with endpoints at A. Label point B on one ray and point C on the other ray. / Point R is the same distance from point E as it is from point F ) one because the arc was made with Step 2 Construct a ray parallel to ACthrough point B. E F compass opening. G Point G is the same distance from point E as it is from point F the same because both arcs were made with B * ) compass opening. C ) Step 3 Construct a ray parallel to AB through point C. ) All rights reserved. A All rights reserved. This means that RG intersectsline ᐉ at the * ) midpoint of perpendicular bisector that RG is the Quick Check. * ) of * ) , and EF EF . * ) * ) 3. a. Use a straightedge to draw EF . Construct FG so that FG EF at point F. Step 4 Label point D where the ray parallel to AC intersects the ray ) parallel to AB . Quadrilateral ABDC has both pairs ofopposite sides parallel. D B Quick Check. 2. Draw two segments. Label their lengths c and d. Construct a quadrilateral with one pair of parallel sides of lengths c and 2d. d c d d F E * ) * ) * ) All rights reserved. C © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc.,publishing as Pearson Prentice Hall. G A * ) * ) b. Draw a line CX and a point Z not on CX . Construct ZB so that ZB CX . Z C c X B Geometry Lesson 3-8 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 3-8 71 Name_____________________________________Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 4-1 3 Finding Congruent Triangles Can you conclude that ABC CDE? Congruent Figures E Listcorresponding vertices in the same order. Lesson Objective 1 Recognize congruent figures and their corresponding parts NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties If ABC CDE, then BAC The diagram above shows BAC The statement ABC CDE LocalStandards: ____________________________________ Notice that BC Vocabulary and Key Concepts. Also, CBA two angles of another triangle, then the third angles B C E F C F Congruent polygons are polygons that have corresponding D sides congruent and corresponding angles congruent. 1Naming Congruent Parts ABC QTJ. List the Q T B congruent corresponding parts. List the corresponding sides and angles in the same order. Angles: A Q B T Sides: AB > QT BC TJ C J AC QJ C A J 2 Using Congruency XYZ KLM, mY 67, and mM 48. Find mX. Use the Triangle Angle-Sum Theorem andthe definition of congruent polygons to find mX. mX mY mZ 180 Triangle Angle-Sum Theorem mZ m⬔M Corresponding angles of congruent triangles are congruent. mZ 48 Substitute 48 for m⬔M. 48 115 180 Substitute. 180 Simplify. mX 65 Geometry Lesson 4-1 Geometry 4 D A EC . . ACB . ABC EDC .Quick Check. WY O MK lWSY O lMVK WS O MV lSWY O lVMK YS O KV lWYS O lMKV 2. It is given that WYS MKV. If mY 35, what is mK? Explain. GHI © Pearson Education, Inc., publishing as Pearson Prentice Hall. DEF Examples. 18 DEC are congruent. The correct way to state this is Angles: G F E72 DE , and AC and BAC 3 true. C Since all of the corresponding sides and angles are congruent, the triangles Sides: I mX , not DCE. is not 3 4 1. WYS MKV. List the congruent corresponding parts. Use three letters for each angle. H © Pearson Education, Inc., publishing as Pearson Prentice Hall. arecongruent. All rights reserved. All rights reserved. D A If two angles of one triangle are congruent to 67 . DEC Using Theorem 4-1, you can conclude that ECD Theorem 4–1 mX DC , BA CDE DCE B mK 35 Corresponding angles of congruent triangles are congruent and have the same measure. 3. Canyou conclude that JKL MNL? Justify your answer. N J L K M No. The corresponding sides are not necessarily equal. Subtract 115 from each side. Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 4-1 73 All-In-One Answers Version A © Pearson Education, Inc., publishing as PearsonPrentice Hall. 70 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date________________ G Triangle Congruence by SSS and SAS Lesson 4-2 Example. 4 Proving Triangles Congruent Lesson Objective Given: CG DG, CN DN, C D, GN # CD Prove: CNG DNG 1 NAEP 2005 Strand: Geometry Prove two triangles congruent using the SSS and SAS Postulates Topic:Transformation of Shapes and Preservation of Properties Local Standards: ____________________________________ C Key Concepts. 1. Given 2. Given 3. GN > GN 3. Reflexive Property of Congruence 4. C D 4. Given 5. CNG DNG 6. 5. ⬔CGN 艑 ⬔DGN Right angles are congruent All rightsreserved. Reasons 1. CG > DG 2. CN > DN All rights reserved. Statements D N 7. Definition of Prove: AEB DCB All rights reserved. C B A D E Statements Reasons 1. ⬔A 艑 ⬔D; ⬔E 艑 ⬔C 1. Given 2. ⬔ABE 艑 ⬔DBC 2. Vertical angles are congruent. © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4. Given: A D, E C, AE DC, EB CB, BA BD F PQR BCA R F D B E 3. AE O CD; AB O BD; EB O BC 3. Given 4. 䉭AEB 艑 䉭DCB 4. Definition of Congruent Triangles Geometry Lesson 4-1 C FDE A Examples. A 1 ProvingTriangles Congruent Given: M is the midpoint of XY , AX > AY Prove: AMX AMY Write a paragraph proof. You are given that M is the midpoint of XY , and AX AY . Midpoint M implies that MX MY . AM > AM by the X SSS Postulate . Daily Notetaking Guide Daily Notetaking Guide Y M , so ReflexiveProperty of Congruence AMX AMY by the Geometry Lesson 4-2 75 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date________________ 2 Using SAS AD > BC . What other information do you Lesson 4-3 B A need to prove ADC BCD by SAS? Reflexive Property of Congruence . ⬔ADC ⬵ ⬔BCD SAS Local Standards: ____________________________________ . Key Concepts. R 3 Are the Triangles Congruent?From the information given, Postulate 4-3: Angle-Side-Angle (ASA) Postulate can you prove RSG RSH? Explain. Given: RSG RSH, SG > SH It is given that RSG RSH and SG > SH. RS RS by the . Reflexive Property of Congruence H G Two pairs of corresponding sides and their included angles arecongruent, so RSG RSH by the SAS All rights reserved. ADC BCD by NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties , you can prove All rights reserved. you know C D You now have two pairs of corresponding congruent sides. Therefore, if TriangleCongruence by ASA and AAS Lesson Objective 1 Prove two triangles congruent using the ASA Postulate and the AAS Theorem It is given that AD > BC. Also, DC > CD by the If two angles and the included side of one triangle are congruent P to two angles and the included side of another triangle, thenthe two triangles are congruent S HGB B K . G NKP N H Postulate. F Theorem 4-2: Angle-Angle-Side (AAS) Theorem J Quick Check. C If two angles and a nonincluded side of one triangle are congruent 1. Given: HF HJ, FG JK, H is the midpoint of GK. to two angles and the corresponding nonincludedside of another Prove: FGH JKH Statements H K Reasons 1. HF > HJ, FG > JK 1. Given 2. H is the midpoint of GK. 2. Given 3. GH > HK 3. Definition of Midpoint 4. 䉭FGH ⬵ 䉭JKH 4. SSS Postulate 2. What other information do you need to prove ABC CDA by SAS? ⬔DCA 艑 ⬔BAC A 8 3. From theinformation given, can you prove AEB DBC? Explain. Given: EB > CB , AE > DB No; you don’t know that ⬔AEB ⬵ ⬔DBC or that AB O DC . B 6 D 7 7 C A E B C D the triangles are congruent triangle, then © Pearson Education, Inc., publishing as Pearson Prentice Hall. G © Pearson Education, Inc.,publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. GHF are congruent. Quick Check. 74 Q P Postulate 4-2: Side-Angle-Side (SAS) Postulate If two sides and the included angle of one triangle are congruent to two sides and the included angle of anothertriangle, then the two triangles triangles congruent H congruent. 6. If two angles of one triangle are congruent to two angles of another triangle, then the third angles are (Theorem 4-1 ). congruent 7. CNG DNG Postulate 4-1: Side-Side-Side (SSS) Postulate If the three sides of one triangle are congruent tothe three sides of another G triangle, then the two triangles are CDM M . G Example. C and I is not congruent to C. Name the triangles that are congruent by the ASA Postulate. The diagram shows N A D and FN > CA > GD. X D O 1 Using ASA Suppose that F is congruent to C F G T N A I If F C, then FC G . Therefore, FNI CAT GDO by T D XGT ASA . Quick Check. 1. Using only the information in the diagram, can you conclude that INF is congruent to either of the other two triangles? Explain. No; Only one angle and one side are shown to be congruent. At least one more congruent side or angle isnecessary to prove congruence with SAS, ASA, or AAS. 76 Geometry Lesson 4-2 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 4-3 Geometry 77 19 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Examples. Quick Check. 2. Write a proof. A Given: /A > /B, AP> BP Prove: APX BPY P It is given that A B and AP > BP. APX BPY by B A Given: B D, AB6CD Prove: ABC CDA Because AB6CD, BAC DCA Angles Theorem. Then ABC congruent. 䉭 NML 艑 䉭 NPO by ASA. B . 3 Planning a Proof Write a plan for a proof that uses AAS. 3. Write a flow proof. P CDAAC , so ABC CDA by C D SRP ⬔QRP R Given: S Q, RP bisects SRQ. Prove: SRP QRP . 4 Writing a Proof Write a two-column proof that uses AAS. Definition of angle bisector S Q B A RP Reasons 1. B D, AB6CD 1. Given 2. lBAC O lDCA 2. If lines are parallel, then alternate interior 2. angles arecongruent. 3. AC > AC 3. Reflexive Property of Congruence 4. ABC CDA 4. AAS Theorem Geometry Lesson 4-3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. C D Statements © Pearson Education, Inc., publishing as Pearson Prentice Hall. Given Given: B D, AB6CD Prove: ABC CDA78 Q S if a pair of AAS RP bisects SRQ Given by the Alternate Interior corresponding sides are congruent. By the Reflexive Property, AC O ⬔MNL 艑 ⬔PNO because vertical angles are All rights reserved. ASA M It is given that NM O NP and ⬔M 艑 ⬔P. X two pairs of corresponding angles and theirincluded sides are congruent, APX BPY by P N RP Reflexive Property of Congruence kSRP O kQRP All rights reserved. Theorem. Because Vertical Angles All rights reserved. the L Given: NM > NP, M P Prove: NML NPO Y AAS 4. Recall Example 4. Explain how you could prove ABC CDA using ASA. Itis given that lB O lD. You know that lBAC O lDCA by the Alternate Interior Angles Theorem. By Theorem 4-1, lBCA O lDAC . By the Reflexive Property of Congruence, AC O AC . Therefore, kABC O kCDA by ASA. Daily Notetaking Guide Daily Notetaking Guide 79 Geometry Lesson 4-3

Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 4-4 2 Using Right Triangles According to legend, one ofNapoleon’s followers used Using Congruent Triangles: CPCTC Lesson Objective 1 Use triangle congruence and CPCTC to prove that parts of two triangles are congruent congruent triangles to estimate the width of a river. On the riverbank, the officer stood up straight and lowered the visor of his cap untilthe farthest thing he could see was the edge of the opposite bank. He then turned and noted the spot on his side of the river that was in line with his eye and the tip of his visor. NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties Local Standards:____________________________________ D P arts are C ongruent All rights reserved. of C ongruent C orresponding . Examples. All rights reserved. CPCTC stands for: T riangles What congruence statements besides 3 4 can you prove from the diagram, in which SL > SR and 1 2 are given? and LSCRSC by SAS . 3 4 because corresponding parts of congruent triangles are congruent. When two triangles are congruent, you can form congruence statements about three pairs of corresponding angles and three pairs of corresponding sides. List the congruence statements. 12 Stretcher S Shaft Sides: SL> SR Given SC SC Reflexive Property of Congruence CL CR Other congruence statement Angles: 1 2 Given 3 4 Corresponding Parts of Congruent Triangles lCLS O lCRS Other congruence statement right angles . DEG and DEF are the angles that the officer makes with the ground. R 6 L 5 Given: DEGand DEF are right angles; EDG EDF. The given states that DEG and DEF are What conditions must hold for that to be true? Rib So the officer must stand © Pearson Education, Inc., publishing as Pearson Prentice Hall. by the Reflexive Property of Congruence, © Pearson Education, Inc., publishing asPearson Prentice Hall. SC C 34 E The officer then paced off the distance to this spot and declared that distance to be the width of the river! 1 Congruence Statements The diagram shows the frame of an umbrella. SC G F Vocabulary. ground must be perpendicular level or flat to the ground, and the .Quick Check. 1. Given: Q R, QPS RSP Prove: SQ > PR P It is given that ⬔Q 艑 ⬔R, ⬔QPS 艑 ⬔RSP. PS O SP by the Reflexive Property of 艑. kQPS O kRSP by ASA. SQ O PR by CPCTC. R Q S 2. Recall Example 2. About how wide was the river if the officer paced off 20 paces and each pace wasabout 212 feet long? The congruence statements that remain to be proved are 50 feet lCLS O lCRS and CL O CR. 80 20 Geometry Lesson 4-4 Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 4-4 81 All-In-One Answers Version A © Pearson Education, Inc., publishing asPearson Prentice Hall. 2 Writing a Proof Using ASA Write a paragraph proof. Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 4-5Name_____________________________________ Class____________________________ Date ________________ The legs of an isosceles triangle Isosceles and Equilateral Triangles The vertex angle of an isosceles triangle is formed by the two are the congruent sides. Lesson Objective 1 NAEP2005 Strand: Geometry Use and apply properties of isosceles triangles Vertex Topic: Relationships Among Geometric Figures Local Standards: ____________________________________ Vocabulary and Key Concepts. The base of an isosceles triangle Legs is the third, or non-congruent, Base side.Theorem 4-3: Isosceles Triangle Theorem B A Theorem 4-4: Converse of Isosceles Triangle Theorem C All rights reserved. A B All rights reserved. the angles opposite those sides are congruent. Example. 1 Using the Isosceles Triangle Theorems Explain why ABC is isosceles. X ABC and XAB are *AB )ABC ← → . Because XA 6 BC , . B the Transitive Property of Congruence Theorem 4-5 You can use C A bisects AB. CD D B Corollary to Theorem 4-3 Y If a triangle is equilateral, then the triangle is equiangular. X Y Z Z X Corollary to Theorem 4-4 Y If a triangle is equiangular, then the triangle isequilateral. XY YZ Z X ZX © Pearson Education, Inc., publishing as Pearson Prentice Hall. and AB © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. CD ' A corollary is a statement that follows immediately from a theorem. 82 Geometry Lesson 4-5Name_____________________________________ Class____________________________ Date________________ 2y 90 Therefore, x 90 Triangle-Sum Theorem 180 Substitute. 180 Simplify. 90 Subtract y 45 Divide each side by 2 . and y 45 90 a regular hexagon. Suppose that a segment is drawnbetween the endpoints of the angle marked x. Find the angle measures of the triangle that is formed. The measure of x is 120. Because the garden is a regular hexagon, the sides have x of the angle measures of a triangle is xyy 180 2y 180 2y 60 y 30 So the angle measures of the triangle are flagstonewalk 120 , 30 , and . 30 Quick Check. 2. Suppose mL 43. Find the values of x and y. M x 90, y 47 3. Use the diagram from Example 3. The path around the garden is made up of rectangles and equilateral triangles. What is the measure of z, the angle at the outside corner of the path? y N 63 x O L x 90Definition of perpendicular mN mL Isosceles Triangle Theorem y x O L Geometry Lesson 4-5 N 83 Congruence in Right Triangles NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties Geometry Lesson 4-5 All-In-One Answers Version A Theorem 4-6:Hypotenuse-Leg (HL) Theorem If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the triangles are congruent. Hypotenuse Legs Examples. by the HL Theorem” for the diagram. Is the student correct? Explain. The diagram shows thefollowing congruent parts. CA > MA CPA lMPA PA PA C hypotenuse Since AC is the of right triangle CPA, and MA is the leg and PA is a hypotenuse P M leg and PA is a of right triangle MPA, the triangles are congruent by HL Theorem The student Daily Notetaking Guide A 1 Proving Triangles CongruentOne student wrote “CPA MPA the 150 84 The bisector of the vertex angle of an isosceles triangle is the perpendicular bisector of the base. right angle. Label each unknown angle y: 120 MO ' LN railroad tie steps z . 180 M Examples. 2 Using Algebra Suppose that mL y. Find the values of x and y. The legsof a right triangle are the two shortest sides, or the sides that are not opposite the y . The measure of the angle marked x is 120, and the sum © Pearson Education, Inc., publishing as Pearson Prentice Hall. congruent S The hypotenuse of a right triangle is its longest side, or the side opposite the rightangle. raised bed garden . By the Isosceles Triangle Theorem, the unknown angles are V Vocabulary and Key Concepts. . isosceles W R from each side. 3 Using Isosceles Triangles The garden shown at the right is in the shape of equal length, so the triangle is No; neither lRUV nor lRVU can be shown tobe congruent to lR. Local Standards: ____________________________________ 2y All rights reserved. All rights reserved. 90 1. Can you deduce that RUV is isosceles? Explain. Lesson Objective 1 Prove triangles congruent using the HL Theorem © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. yy T U Quick Check. Lesson 4-6 Transitive Property of Equality mN mNMO mMON 180 to AC . By the definition of an isosceles triangle, ABC is isosceles. Name_____________________________________Class____________________________ Date ________________ Given mN y C . ACB Daily Notetaking Guide Daily Notetaking Guide mL y , ABC the Converse of the Isosceles Triangle Theorem conclude that AB > The bisector of the vertex angle of an isosceles triangle is the of the base. ← → Thediagram shows that XAB ACB. By B A BC XAB A angles formed alternate interior by XA , BC , and the transversal the sides opposite those sides are congruent. perpendicular bisector and the legs. Base Angles ← → ← → If two angles of a triangle are congruent, then The base angles of an isoscelestriangle are formed by the base C If two sides of a triangle are congruent, then AC congruent sides (legs). Angle is Daily Notetaking Guide . correct. Geometry Lesson 4-6 Geometry 85 21 Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ 2 Flow Proof—Using the HL Theorem XYZ is isosceles. From vertex X, Quick Check. a perpendicular is drawn to YZ,intersecting YZ at point M. Explain why XMY XMZ. 1. R L 3 S O 3 X 3 M 5 5 5 Q P Z XMY and XMZ are right angles right triangles Definition of perpendicular lines Definition of right triangle 2. Write a paragraph proof of Example 2. XMY XMZ XY XZ Definition of HL Theorem 3 Two-Column Proof—Usingthe HL Theorem C D Given: ABC and DCB are right angles, AC > DB Prove: ABC DCB Statements E Reasons 1. ABC and DCB are B A 1. Given right angles 2. ABC and DCB 2. Definition of Right Triangle 3. Given are right triangles 3. AC > DB BC O CB 4. 5. ABC 86 4. Reflexive Property of CongruenceDCB 5. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Reflexive Property of Congruence © Pearson Education, Inc., publishing as Pearson Prentice Hall. XM XM isosceles 䉭LMN 艑 䉭OQP It is given that XM is perpendicular to YZ . This means that lYMX and ⬔ZMX are right angles,since that is the definition of perpendicular lines. It follows, then, that 䉭YMX and 䉭ZMX are right triangles, since they each contain a right angle. It is given that 䉭XYZ is isosceles, so XY O XZ by the definition of isosceles. XM O XM by the Reflexive Property of Congruence. Therefore, since thehypotenuses are congruent and one leg is congruent, 䉭XMY 艑 䉭XMZ by the HL Theorem. 3. You know that two legs of one right triangle are congruent to two legs of another right triangle. Explain how to prove the triangles are congruent. Since all right angles are congruent, the triangles are congruentby SAS. If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the triangles are congruent (HL Theorem). Geometry Lesson 4-6 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________Class____________________________ Date________________ Lesson Objectives 1 Identify congruent overlapping triangles 2 Prove two triangles congruent by first proving two other triangles congruent Examples. 3 Using Two Pairs of Triangles Write a paragraph proof. NAEP 2005 Strand: GeometryTopic: Relationships Among Geometric Figures Plan: XPW YPZ by AAS if WXZ SAS congruent by Examples. and HG F FG E D , respectively. 2 Using Common Parts Write a plan for a proof that Y Z does not use overlapping triangles. All rights reserved. These parts are W are congruent. By the AngleAddition Postulate, ZWM YXM. So ZWM YXM by ASA . . Quick Check. 1. The diagram shows triangles from the scaffolding that workers used when they repaired and cleaned the Statue of Liberty. a. Name the common side in ACD and BCD. CD A B F C D E © Pearson Education, Inc., publishing asPearson Prentice Hall. . © Pearson Education, Inc., publishing as Pearson Prentice Hall. vertical angles . CPCTC SAS congruence holds by . Converse of the Isosceles Triangle Theorem CPCTC AAS YPZ are congruent. Therefore, vertical angles XPW YPZ by Plan: CBE CDA by Look at MWX. MW > MXby the ZMW YMX because because by CPCTC, and XPW ZYW C Write a two-column proof to show that CBE CDA. X You can prove these triangles congruent using ASA as follows: So ZW > YX by . Therefore, XWZ YZW by Reflexive Property of Congruence SAS. WXZ Z W Given: CA > CE, BA > DELabel point M where ZX intersects WY. 䉭ZWM 艑 䉭YXM P . 4 Separating Overlapping Triangles M Given: ZXW YWX, ZWX YXW Prove: ZW > YX ZW > YX by CPCTC if . These angles , XWZ YZW. WZ > ZW by the right angles All rights reserved. EG lZYW Proof: You are given XW > YZ. Because XWZand YZW are G H are shared by DFG and EHG. Parts of sides DG and Y if XWZ YZW. These triangles are CPCTC are congruent by DFG and EHG share. Identify the overlapping triangles. X Given: XW > YZ, XWZ and YZW are right angles. Prove: XPW YPZ Local Standards:____________________________________ 1 Identifying Common Parts Name the parts of the sides that 87 Name_____________________________________ Class____________________________ Date ________________ Using Corresponding Parts of Congruent Triangles Lesson 4-7 GeometryLesson 4-6 B if CBE CDA. This if CB > CD. X A D E Proof: Statement Reason 1. BCE DCA 1. Reflexive Property of Congruence 2. CA CE , BA DE 2. Given 3. CA CE , BA DE 3. Congruent sides have equal measure. 4. CA BA CE DE 4. Subtraction Property of Equality 5. CA BA CB 5. Segment AdditionPostulate CE DE CD 6. CB CD 6. Substitution 7. CB > CD 7. Definition of Congruence 8. CBE CDA 8. SAS 9. CBE CDA 9. CPCTC b. Name another pair of triangles that share a common side. Name the common side. Answers may vary. Sample: 䉭ABD and 䉭CBD; BD 88 22 Geometry Lesson 4-7Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 4-7 89 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. Given T Which two triangles are congruent by the HL Theorem? Write a correct congruence statement. All rights reserved.XMY and XMZ are XM ⊥ YZ All rights reserved. M All rights reserved. N Y Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date________________ Quick Check. Lesson 5-1 A 2. Write a paragraph proof. Given: ACD BDC Prove: CE > DE B Midsegments of Triangles Lesson Objective E 1 C D NAEP 2005 Strand: Geometry Use properties of midsegments to solve problems Topics:Relationships Among Geometric Figures Local Standards: ____________________________________ It is given that 䉭ACD 艑 䉭BDC, so lADC O lBCD by CPCTC. This means that 䉭CED is isosceles by the Converse of Isosceles Triangle Theorem. Then CE O DE by the definition of isoscelestriangle. Vocabulary and Key Concepts. Q 3. Write a two-column proof. Given: PS > RS, PSQ RSQ Prove: QPT QRT P R T All rights reserved. All rights reserved. Theorem 5-1: Triangle Midsegment Theorem S Given 2. QS O QS 2. Reflexive Property of Congruence 3. 䉭PSQ 艑 䉭RSQ 3. SAS 4. PQ ORQ 4. CPCTC 5. lPQT O lRQT 5. CPCTC 6. QT O QT 6. Reflexive Property of Congruence 7. 䉭QPT 艑 䉭QRT 7. SAS Examples. A D F E Reason Given 2. AD O AD 2. Reflexive Property of Congruence 3. 䉭ACD 艑 䉭AED 3. AAS 4. CD O ED 4. CPCTC 5. lBDC O lFDE 5. Vertical Angles are Congruent6. 䉭BDC 艑 䉭FDE 6. ASA 7. BD O FD 7. CPCTC Geometry Lesson 4-7 © Pearson Education, Inc., publishing as Pearson Prentice Hall. C B 1. 90 2 Identifying Parallel Segments Find mAMN and mANM. MN and BC Triangle Midsegment Theorem, so AMN B by the Corresponding Angles Postulate. 75m⬔AMN 22 P Definition of perimeter 24 NP 22 NP 46 24 Substitute 60 60 Simplify. NP 14 Subtract 22 24 for MN and Y for MP. 46 N Z from each side. Use the Triangle Midsegment Theorem to find YZ. 1 MP YZ Triangle Midsegment Theorem 2 1 22 44 YZ YZ 2 Substitute 22 Multiply each side by forMP. 2 . Geometry Lesson 5-1 91 A E M N B D by the Theorem. mANM Isosceles Triangle 60 1. AB 10 and CD 18. Find EB, BC, and AC. because congruent angles have the same measure. In AMN, AM AN , so mANM M NP MN MP Quick Check. A corresponding angles. MN6BC by the Because theperimeter of MNP is 60, you can find NP. Name_____________________________________ Class____________________________ Date ________________ 75 C 75 C EB 9; BC 10; AC 20 B to swim the length of the lake, as shown in the photo. Explain how Dean could use the Triangle MidsegmentTheorem to measure the length of the lake. To find the length of the lake, Dean starts at the middle of the left edge of the lake and paces straight along that end of the lake. He counts the number of strides (35). Where the lake ends, he sets a stake. He paces the same number of strides (35) in the samedirection and sets another stake. The first stake marks the midpoint of one side of a triangle. Dean paces from the second stake straight to the middle of the other end of the lake. He counts the number of his strides (236). 65 Y 118 strides. 236 35 35 From this midpoint of the second side of the triangle, ?he returns to the midpoint of the first side, counting the number of strides (128). Dean has paced a triangle. He has also formed a midsegment whose third side is the By the length of a triangle of the lake. Triangle Midsegment Theorem, the segment connecting the two midpoints is half the distance acrossthe lake. So, Dean multiplies the length of the midsegment by 2 to find U Z V 65; UV6XY so ⬔VUZ and ⬔YXZ are corresponding and congruent. 118 128 © Pearson Education, Inc., publishing as Pearson Prentice Hall. the number of strides back half toward the second stake. He paces 2. Critical ThinkingFind mVUZ. Justify your answer. X Dean finds the midsegment of the second side by pacing exactly All rights reserved. All rights reserved. by substitution. 3 Applying the Triangle Midsegment Theorem Dean plans © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc.,publishing as Pearson Prentice Hall. are cut by transversal AB, so AMN and B are The perimeter of MNP is 60. Find NP and YZ. Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ mAMNits length. A midsegment of a triangle is a segment connecting the midpoints of two sides. 1 Finding Lengths In XYZ, M, N, and P are midpoints. lCAD O lEAD, lC O lE 1. half to the third side, and is to prove a theorem. 1. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.Reason PS O RS, lPSQ O lRSQ Statement parallel is A coordinate proof is a form of proof in which coordinate geometry and algebra are used Statement 1. 4. Write a two-column proof. Given: CAD EAD, C E Prove: BD > FD If a segment joins the midpoints of two sides of a triangle, then the segment 3.CD is a new bridge being built over a lake as shown. Find the length of the bridge. 963 ft 2640 ft C Bridge 963 ft D 1320 ft ? the length of the lake. 92 Geometry Lesson 5-1 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 5-1 Geometry 93 23 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 5-2 Examples. Bisectors in Triangles Lesson Objective 1 Name_____________________________________Class____________________________ Date ________________ 1 Applying the Perpendicular Bisector Theorem Use the map of NAEP 2005 Strand: Geometry Use properties of perpendicular bisectors and angle bisectors Washington, D.C. Describe the set of points that are equidistant from the LincolnMemorial and the Capitol. Topics: Relationships Among Geometric Figures Local Standards: ____________________________________ Vocabulary and Key Concepts. All rights reserved. If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.Theorem 5-3: Converse of the Perpendicular Bisector Theorem If a point is equidistant from the endpoints of a segment, then it is All rights reserved. Theorem 5-2: Perpendicular Bisector Theorem on the perpendicular bisector of the segment. Converse of the Perpendicular Bisector The Theorem states:If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment. If a point is on the bisector of an angle, then it is segment from the point to the line. * ) * ) D is 3 in. from AB and AC 0 1 2 3 4 5 6 94 . C B Geometry Lesson 5-2Name_____________________________________ Class____________________________ Date ________________ * 2 Using the Angle Bisector Theorem C Find x, FB, and FD in the diagram at right. FD FB Theorem 7x 37 2x 5 7x 2x 5x B Angle Bisector 42 Add 42 37 Subtract x 8.4 Substitute. FD 7(8.4 ) 37 21.8 Substitute. 2x 1. Use the information given in the diagram. CD is the perpendicular C bisector of AB. Find CA and DB. Explain your reasoning. * ) CA 5; DB 6; CD is the perpendicular bisector of AB; therefore CA CB and DA DB. 5 . Geometry Lesson 5-2 6 ) 95 Concurrent Lines, Medians, andAltitudes Lesson Objective 1 Identify properties of perpendicular bisectors and angle bisectors 2 Identify properties of medians and altitudes of a triangle B D NAEP 2005 Strand: Geometry Topics: Relationships Among Geometric Figures Local Standards: ____________________________________Vocabulary and Key Concepts. 2. a. According to the diagram, how far is K from EH ? From ED ? 2x K C 10 E (x 20) All rights reserved. D 10; 10 All rights reserved. ) E Name_____________________________________ Class____________________________ Date________________ 5 A 7x 37 fromeach side. Lesson 5-3 ) F to each side. Divide each side by FB 2( 8.4 ) 5 21.8 D 2x 5 A Substitute. Daily Notetaking Guide Daily Notetaking Guide Quick Check. of the segment whose endpoints are the Lincoln Memorial and the Capitol. H Theorem 5-6 The perpendicular bisectors of the sides of a triangleare concurrent at a point equidistant from the vertices. Theorem 5-7 The bisectors of the angles of a triangle are ) b. What can you conclude about EK ? concurrent at a point equidistant from the sides. ) EK is the angle bisector of lDEH. Theorem 5-8 The medians of a triangle are concurrent at a point thatis two-thirds the distance from c. Find the value of x. 20 d. Find mDEH. 80 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. each vertex to the midpoint of the opposite side. DC 2 DJ 3 EC 2 EG 3 FC D 2 H FH 3 G Theorem 5-9 The lines that contain the altitudes of a triangle are concurrent. C E J F Concurrent lines are three or more lines that meet in one point. The point of concurrency is the point at which concurrent lines intersect. A circle is circumscribed about a polygon when the vertices Circumscribed of the polygon areon the circle. S The circumcenter of a triangle is the point of concurrency of circle QC SC RC the perpendicular bisectors of a triangle. Median C A median of a triangle is a segment that has as its endpoints a vertex of the Q R triangle and the midpoint of the opposite side. 96 24 Geometry Lesson 5-2Geometry Daily Notetaking Guide Daily Notetaking Guide circumcenter Geometry Lesson 5-3 97 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. The distance from a point to a line is the length of the perpendicular © Pearson Education, Inc., publishing asPearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. angle, then it is on the angle bisector. D from the Lincoln Memorial and perpendicular bisector the Capitol must be on the If a point in the interior of an angle is equidistant from the sides of the A equidistant A point thatis equidistant from the sides of the angle. Theorem 5-5: Converse of the Angle Bisector Theorem All rights reserved. Theorem 5-4: Angle Bisector Theorem Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________Date ________________ 101 of a triangle are bisectors of the angles incenter concurrent at a point Park rd V va vertex to the line containing the opposite side. Z le ou Theorem 5-7 states that the U B The roads form a triangle around the park. I An altitude of a triangle is the perpendicular segment from aa os ip ar M X bisectors of the triangle. fountain equidistant from three straight roads that enclose a park. Explain how they can find the location. XI YI ZI Y T The incenter of a triangle is the point of currency of the angle 2 Applying Theorem 5-7 City planners want to locate a circle Inscribed tangent to thecircle. Hig hw ay A circle is inscribed in a polygon if the sides of the polygon are Name_____________________________________ Class____________________________ Date ________________ Andover Road from the sides. equidistant inside a side . Obtuse Triangle Altitude is outside . . All rightsreserved. Right Triangle Altitude is Acute Triangle Altitude is All rights reserved. The city planners should find the point of concurrency of the of the triangle formed angle bisectors by the three roads and locate the fountain there. 3 Finding Lengths of Medians M is the centroid of WOR, and WM 16. W FindWX. The centroid of a triangle is the point of intersection of the medians of the triangle. centroid The The medians of a triangle are concurrent at a point that is (1, 7) , XY lies on the vertical line x 1 . The perpendicular bisector of XY is the horizontal line that 7 passes through (1, 1 2 ) or (1, 4) (1, 1) (5, 1) 6(3, 4) 4 , so the 2 equation of the perpendicular bisector of XY is y 4 . Because X has coordinates y 8Y and Y has X Z O and Z has 2 4 6 x , XZ lies on the horizontal line y 1 . The perpendicular bisector of XZ is the vertical line that (1 2 5, 1) or (3, 1) , so the equation © Pearson Education, Inc., publishingas Pearson Prentice Hall. (1, 1) Because X has coordinates © Pearson Education, Inc., publishing as Pearson Prentice Hall. circumscribes XYZ. All rights reserved. Y R X of the opposite side. (Theorem 5-8) 1 Finding the Circumcenter Find the center of the circle that passes through O the distance fromeach vertex to the midpoint two-thirds Examples. coordinates Z M of the triangle. coordinates is the point of concurrency of the medians of a triangle. The orthocenter of a triangle is the point of intersection of the lines containing the altitudes Because M is the WM 16 24 2WX 3 of WOR, WM centroid . 2WX Theorem WX Substitute 5-8 3 2 16 for WM. 3 WX Multiply each side by 3 2 . of the perpendicular bisector of XZ is x 3 . Draw the lines y 4 and x 3 . They intersect at the point (3, 4) . This point is the center of the circle that circumscribes XYZ. 98 Geometry Lesson 5-3 Daily Notetaking Guide DailyNotetaking Guide Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ y Lesson Objectives 5 1 Write the negation ofa statement and the inverse and contrapositive of a conditional statement 2 Use indirect reasoning NAEP 2005 Strand: Geometry Topics: Mathematical Reasoning Local Standards: ____________________________________ (4, 3) Vocabulary and Key Concepts. 5 5 All rights reserved. All rightsreserved. 5x b. Critical Thinking In Example 1, explain why it is not necessary to find the third perpendicular bisector. Theorem 5-6: All the perpendicular bisectors of the sides of a triangle are concurrent. 2. a. The towns of Adamsville, Brooksville, and Cartersville want to build a library that is equidistantfrom the three towns. Show on the diagram where they should build the library. Explain. Brooksville Cartersville b. What theorem did you use to find the location? The perpendicular bisectors of the sides of a triangle are concurrent at a point equidistant from the vertices (Theorem 5-6). K 8 All-In-OneAnswers Version A If an angle is a straight angle, then its measure is 180. Symbolic Form Negation (of p) An angle is not a straight angle. p Not Inverse If an angle is not a straight angle, then its measure is not 180. p S q If not p, then not q. Contrapositive If an angle’s measure is not 180, then it is not astraight angle. q S p If not q, then not p. pSq You Read It If p, then q. p. Step 1 State as an assumption the opposite (negation) of what you want to prove. Step 2 Show that this assumption leads to a contradiction. Step 3 Conclude that the assumption must be false and that what you want to prove mustbe true. The negation of a statement has the opposite truth value from that of the original statement. The inverse of a conditional statement negates both the hypothesis and the conclusion. The contrapositive of a conditional switches the hypothesis and the conclusion and negates both. In indirectreasoning all possibilities are considered and then all but one are proved false. L Geometry Lesson 5-3 Example Conditional Equivalent statements have the same truth value. median; MY is a segment drawn from vertex M to the midpoint of the opposite side. 100 Statement A conditional and itscontrapositive always have the same truth value. Y 4. Is MY a median, an altitude, or neither? Explain. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Adamsville Draw segments connecting the towns. Build the library at the intersection point of the perpendicular bisectors of the segments.3. Using the diagram in Example 3, find MX. Check that WM MX WX. Negation, Inverse, and Contrapositive Statements Writing an Indirect Proof © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Inverses, Contrapositives, andIndirect Reasoning Lesson 5-4 Quick Check. 1. a. Find the center of the circle that you can circumscribe about the triangle with vertices (0, 0), (8, 0), and (0, 6). 99 Geometry Lesson 5-3 X M Daily Notetaking Guide The remaining possibility must be true. An indirect proof is a proof involving indirectreasoning. Daily Notetaking Guide Geometry Lesson 5-4 Geometry 101 25 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________Class____________________________ Date ________________ 4 Identifying Contradictions Identify the two statements that contradict Examples. each other. I. P, Q, and R are coplanar. II. P, Q, and R are collinear. III. mPQR 60 1 Writing the Negation of a Statement Write the negation of “ABCD is nota convex polygon.” opposite The negation of a statement has the truth value. The not negation of is not in the original statement removes the word . contradict Two statements each other when they cannot both be true at the same time. Examine each pair of statements to see whether they contradict eachother. The negation of “ABCD is not a convex polygon” is All rights reserved. of the conditional statement “If ABC is equilateral, then it is isosceles.” To write the inverse of a conditional, negate both the hypothesis and the conclusion. Hypothesis Conclusion T If ABC is then it is isosceles. Negate both. notequilateral , not isosceles II and III P, Q, and R are coplanar, and mPQR 60. P, Q, and R are collinear, and mPQR 60. Three points that lie on the same line are both coplanar and collinear, so these two statements do not contradict each other. Three points that lie on an angle are coplanar, so these twostatements do not contradict each other. If three distinct points are collinear, they form a straight angle, so mPQR cannot equal 60. Statements II and do III contradict each other. . 5 Indirect Proof Write an indirect proof. Prove: ABC cannot contain two obtuse angles. T T Step 1 Assume as true the Switchand negate both. not isosceles , then it is not equilateral . 3 The First Step of an Indirect Proof Write the first step of an indirect proof. Prove: A triangle cannot contain two right angles. In the first step of an indirect proof, you assume as true the negation of what you want to prove. Because you want toprove that a triangle cannot contain two right angles, can you assume that a triangle contain two right angles. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Conclusion © Pearson Education, Inc., publishing as Pearson Prentice Hall. Hypothesis If ABC is equilateral, then it is isosceles.Contrapositive: If ABC is I and III P, Q, and R are coplanar and collinear. T then it is To write the contrapositive of a conditional, switch the hypothesis and conclusion, then negate both. Conditional: I and II The first step is “Assume that a triangle contains two right angles.” opposite of what you want toprove. That is, assume that ABC contains two Let lA and lB be Step 2 If A and B are obtuse obtuse , mA mB mC obtuse Triangle Angle-Sum This contradicts the which states that mA mB mC Step 3 The assumption in Step 1 must be contain two angles. . false 180 . All rights reserved. T Inverse: T If ABCis equilateral, Conditional: All rights reserved. “ABCD is a convex polygon.” 2 Writing the Inverse and Contrapositive Write the inverse and contrapositive Theorem, 180 . . ABC cannot angles. obtuse Quick Check. 1. Write the negation of each statement. a. mXYZ 70. The measure of ⬔XYZ is not morethan 70. b. Today is not Tuesday. Today is Tuesday. Geometry Lesson 5-4 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________Class____________________________ Date________________ Lesson 5-5 2. Write (a) the inverse and (b) the contrapositive of Maya Angelou’s statement “If you don’t stand for something, you’ll fall for anything.” a. b. 103 Geometry Lesson 5-4 Inequalities in Triangles Lesson Objectives 1 Useinequalities involving angles of triangles 2 Use inequalities involving sides of triangles If you stand for something, you won’t fall for anything. NAEP 2005 Strand: Geometry Topics: Relationships Among Geometric Figures Local Standards: ____________________________________ If you won’t fall foranything, then you stand for something. Key Concepts. All rights reserved. The shoes cost more than $20. Comparison Property of Inequality All rights reserved. 3. You want to prove each statement true. Write the first step of an indirect proof. a. The shoes cost no more than $20. If a b c and c 0, then a b. Corollary to the Triangle Exterior Angle Theorem The measure of an b. mA mB m⬔1 III. FG > KL © Pearson Education, Inc., publishing as Pearson Prentice Hall. II. FG ' KL I and II; Two segments cannot be parallel as well as perpendicular. I and II contradict each other. Two segments can be paralleland congruent. I and III do not contradict each other. Two segments can be perpendicular and congruent. II and III do not contradict each other. 5. Critical Thinking You plan to write an indirect proof showing that X is an obtuse angle. In the first step you assume that X is an acute angle. What have youoverlooked? © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4. Identify the two statements that contradict each other. Explain. I. FG6KL exterior m2 and m⬔1 3 angle of a triangle is greater interior than the measure of each of its remote mlA # mlB angles. 2 m3 Theorem 5-10 If two sidesof a triangle are not congruent, then 1 Y the larger angle lies opposite the longer side. If XZ XY, then mY mZ. X Theorem 5-11 If two angles of a triangle are not congruent, then Z B C the longer side lies opposite the larger angle. If mA mB, then BC AC. A L Theorem 5-12: Triangle Inequality Theorem Thesum of the lengths of any two sides of a triangle is greater than the length of the third side. M LM MN LN MN LN LM N LN LM MN ⬔X could be a right angle. 104 26 Geometry Lesson 5-4 Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 5-5 105 All-In-One Answers Version A ©Pearson Education, Inc., publishing as Pearson Prentice Hall. 102 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________Class____________________________ Date ________________ 4 Finding Possible Side Lengths In FGH, FG 9 m and GH 17 m. Examples. Describe the possible lengths of FH. 1 Applying the Corollary Explain why m4 m5. 4 is an B 5 angle of XYC. The Corollary exterior to the Exterior Angle Theoremstates that the measure greater of an exterior angle of a triangle is 4 X A than the measure of each of its remote interior angles. The Triangle Inequality Theorem: The sum of the lengths of any two sides of greater a triangle is than the length of the third side. Y 2 1 Solve three inequalities. 3 All rightsreserved. angles formed by transversal BC . Because AB6XY, you can conclude that 2 ⬔5 by the Corresponding Angles Postulate. Thus, m2 m⬔5 . Substituting m⬔5 for m2 in the inequality m4 m2 produces the inequality m⬔4 m⬔5 . List the angles from largest to smallest. Theorem 5-10: If two sides ofa triangle are not congruent, then the larger angle lies opposite the longer side. All rights reserved. 12 Y 20 R . The opposite angle, ⬔R , is the smallest. From largest to smallest, the angles are ⬔G , ⬔Y , ⬔R . 3 Using the Triangle Inequality Theorem Can a triangle have sides with the given lengths?Explain. a. 2 cm, 2 cm, 4 cm Triangle Inequality According to the Theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side. 22 ⬐ 4 The sum of 2 and 2 No , a triangle cannot is not greater than 4. can 1. Use the diagram in Example 1. If CY XY, explain why m4m1. A 27 ft C 21 ft 18 ft B 3. Can a triangle have sides with the given lengths? Explain. a. 2 m, 7 m, and 9 m no; 2 7 is not greater than 9 b. 4 yd, 6 yd, and 9 yd yes; 4 6 9, 6 9 4, and 4 9 6 Daily Notetaking Guide Daily Notetaking Guide Lesson 6-1 Examples. Classifying Quadrilaterals Lesson Objective 1Define and classify special types of quadrilaterals 1 Classifying by Coordinate Methods Determine the most precise name NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures for the quadrilateral with vertices Q(4, 4), B(2, 9), H(8, 9), and A(10, 4). Graph quadrilateral QBHA.Local Standards: ____________________________________ Find the slope of each side. 94 Slope of QB 2 (4) 99 Slope of BH 8 2 y B Key Concepts. H 8 Special Quadrilaterals ( 6 Parallelogram Rhombus Trapezoid Isosceles Trapezoid All rights reserved. Rectangle Q All rights reserved. QuadrilateralQuadrilateral of irs es 2 pairs of pa 1 pair of id No llel s parallel sides parallel sides ra pa Square A 4 2 O 2 4 6 Slope of HA 10 5 2 0 25 2 0 . not equal . trapezoid One pair of opposite sides is parallel, so QBHA is a . Next, use the distance formula to see whether any pairs of sides are congruent. kite A is aquadrilateral with two pairs adjacent sides of ______________________ opposite congruent and no ________ sides ________________________ congruent. rectangle A is a parallelogram with right angles four______________________ __________________________. trapezoid A is a quadrilateralwith exactly one pair of ________________________ parallel sides ________________________ ________________________ . 4 4 4 equal BH is parallel to QA because their slopes are square A is a parallelogram with congruent sides ___ four__________________ ______________ and four rightangles ___ ___________________ . isosceles trapezoid An _____________________ ________________________ is a trapezoid whose nonparallel sides are congruent. Geometry Lesson 6-1 All-In-One Answers Version A Daily Notetaking Guide (2 (4))2 (9 4 )2 4 25 "29 (10 8)2 ( 4 9 )2 4 25 "29 10 BH(8 (2))2 ( 9 9 )2 0 100 2 2 QA (4 10 ) ( 4 4 ) 0 196 QB HA © Pearson Education, Inc., publishing as Pearson Prentice Hall. rhombus A is a parallelogram with congruent sides four_____________________ ______________________. 10 8 Slope of QA x 8 ) 4 9 2 HA is not parallel to QB because theirslopes are parallelogram A is a quadrilateral with both pairs of opposite parallel sides ___________________ ________________________ . 107 Geometry Lesson 5-5 Name_____________________________________ Class____________________________ Date ________________ © PearsonEducation, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Quick Check. 15 12 8 Name_____________________________________ Class____________________________ Date________________ 108 FH m. have sides with these lengths.Geometry Lesson 5-5 Kite 26 9 x 15 8 12 15 Yes , a triangle 106 m and shorter than FH 4. A triangle has sides of lengths 3 in. and 12 in. Describe the possible lengths of the third side. have sides with these lengths. b. 8 in., 15 in., 12 in. 8 15 12 8 17 26 ⬔A, ⬔C, ⬔B . The opposite angle, ⬔G , is thelargest. The shortest side is 9 9 FH 8 2. List the angles of ABC in order from smallest to largest. 14 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Find the angle opposite each side. G 12 © Pearson Education, Inc., publishing as Pearson Prentice Hall. No two sides of RGY arecongruent, so the largest angle lies opposite the longest side. 20 FH 17 m⬔1 m⬔3 by the Isosceles Triangle Theorem m⬔4 m⬔3 by the Corollary to the Triangle Exterior Angle Theorem m⬔4 m⬔1 by Substitution 2 Applying Theorem 5-10 In RGY, RG 14, GY 12, and RY 20. The longest side is 8 FHmust be longer than All rights reserved. corresponding 17 FH C The remote interior angles are ⬔2 and ⬔3 , so m4 m2. 2 and 5 are FH 9 Because QB HA an isosceles trapezoid , QBHA is 14 . U 2 Using the Properties of Special Quadrilaterals In parallelogram RSTU, mR 2x 10 and mS 3x 50. Find x. TRSTU is a parallelogram. Given ST6 Definition of parallelogram RU (3x 50) (2 x 10) R S mR mS If lines are parallel, then interior angles on the same side of a transversal are 180 (2x 10) (3x 50) 180 5x 2x 10 for m⬔R and 40 180 Simplify. 5x Subtract 40 from each side. x Daily Notetaking Guide Substitute140 28 supplementary 3x 50 . for m⬔S. Divide each side by 5. Geometry Lesson 6-1 Geometry 109 27 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________Name_____________________________________ Class____________________________ Date ________________ Quick Check. Lesson 6-2 1. a. Graph quadrilateral ABCD with vertices A(3, 3), B(2, 4), C(3, 1), and D(2, 2). b. Classify ABCD in as many ways as possible. y B A ABCD is aquadrilateral because it has four sides. It is a parallelogram because both pairs of opposite sides are parallel. It is a rhombus because all four sides are congruent. It is a rectangle because it has four right angles and its opposite sides are congruent. It is a square because it has four right angles and itssides are all congruent. Properties of Parallelograms Lesson Objectives 1 2 x NAEP 2005 Strand: Geometry Use relationships among sides and among angles of parallelograms Use relationships involving diagonals of parallelograms and transversals Topic: Relationships Among Geometric Figures LocalStandards: ____________________________________ C D c. Which name gives the most information about ABCD? Explain. All rights reserved. All rights reserved. Vocabulary and Key Concepts. Square; the name square means a figure is both a rectangle and a rhombus, since its sides are allcongruent and it has four right angles. Theorem 6-1 C A Opposite angles of a parallelogram are congruent D . R S Theorem 6-3 bisect The diagonals of a parallelogram 4b 2 S N 3a 8 T each other. T BD B A D C transversal. E F DF All rights reserved. 3b 2 © Pearson Education, Inc., publishing asPearson Prentice Hall. L 5a 4 U Theorem 6-4 If three (or more) parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2. Find the values of the variables in the rhombus. Then find thelengths of the sides. a 2, b 4, LN ST NT SL 14 B sides of a parallelogram are congruent. Opposite Theorem 6-2 Consecutive angles of a polygon share a common side. K J L M In JKLM, J and M are consecutive angles, as are J and K . J and L are not consecutive. Examples Geometry Lesson 6-1 DailyNotetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ 1 Using Algebra Find the value of x in ABCD. Then find mA. x 15 135 x 135 2x 120 60 Add x to each side. Subtract 15 60 Divide each side by2 . 15 75 Substitute 60 D for x. . supplementary 180 75 mA C Consecutive angles of a parallelogram are mA mB 180 mA Place a corner of the top of the card on the first line of the lined paper. Place a corner that forms a consecutive angle with the first corner on the sixth line. Mark the points where thelines intersect one side of the card. Mark the points where the lines intersect the opposite side of the card. Connect the marks on opposite sides using a straightedge. (135 x) from each side. x Subtract for mB. 75 Substitute 105 75 from each side. All rights reserved. mB segments on every transversal.Explain how to divide a blank card into five equal rows using Theorem 6-4 and a sheet of lined paper. . All rights reserved. 15 off congruent segments on one transversal, then they cut off congruent (x 15) Opposite angles of a parallelogram are congruent 2xName_____________________________________ Class____________________________ Date ________________ B A 2 Using Algebra Find the values of x and y in KLMN. 2x 5 5y 14y H The diagonals of a parallelogram bisect each other. Substitute for x in 7y 16 the second equation to solve for y. )5 5y 32 5 5y 27 Distribute. 5y 27 Simplify. 9y 3 y Subtract 14y Divide each side by ( ) 16 x7 3 x 5 from each side. 9 . Substitute 3 for y in the first equation to solve for x. © Pearson Education, Inc., publishing as Pearson Prentice Hall. x 7y 16 G M 2x 5 5y 7y 16 (3y 37) (6y 4) E N 14y F A 7y 16 ( 1. Find thevalue of y in EFGH. Then find mE, mF, mG, and mH. L x 2 Quick Check. y 5 11; mlE 5 70, mlF 5 110, mlG 5 70, and mlH 5 110 © Pearson Education, Inc., publishing as Pearson Prentice Hall. K 111 Geometry Lesson 6-2 2. Find the values of a and b. X a a 16, b 14 b 10 W ← → ← → ← → Y 8 2a b 2 Z← → 3. In the figure, DH 6 CG 6 BF 6 AE, AB BC CD 2, and EF 2.5. Find EH. D 7.5 C Simplify. B So x 5 and y 3 . A H 2 G 2 F 2.5 2 E 3 Using Theorem 6-4 Theorem 6-4 states If three (or more) parallel lines cut 112 28 Geometry Lesson 6-2 Geometry Daily Notetaking Guide Daily Notetaking GuideGeometry Lesson 6-2 113 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 110 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________

Name_____________________________________ Class____________________________ Date ________________ Proving That a Quadrilateral Is a Parallelogram Lesson 6-3 2 Is the Quadrilateral a Parallelogram? Can you prove the quadrilateral is a parallelogram from what is given? Explain.Lesson Objective 1 NAEP 2005 Strand: Geometry Determine whether a quadrilateral is a parallelogram a. Given: AB > CD Prove: ABCD is a parallelogram. Topic: Geometry Local Standards: ____________________________________ All you know about the quadrilateral is that only one pair of oppositesides is . congruent Therefore, you cannot conclude that the quadrilateral is a parallelogram. Key Concepts. the quadrilateral is a parallelogram. of a quadrilateral are congruent, Theorem 6-7 If the diagonals of a quadrilateral then the quadrilateral is a parallelogram. each other, bisect Theorem 6-8 If onepair of opposite sides of a quadrilateral is both parallel and B 8x 12 y9 If the diagonals of quadrilateral ABCD bisect each other, Theorem 6-7 then ABCD is a parallelogram by . Write and solve two equations to find values of x and y for which the diagonals bisect each other. 2y 80 G 10x 24 C D Diagonalsof parallelograms bisect each other. 10x 24 8x 12 2x If x 114 24 12 Collect the variables on one side. 2x 36 Solve. x 18 18 and y 2y 80 y9 F The sum of the measures of the angles of a polygon is (n 2)180 where n represents the number of sides, so the sum of the measures of the angles of a 360quadrilateral is 4 2 180 . 105 G 75 ) E 75 H congruent, then the quadrilateral is a parallelogram. Because both pairs of opposite angles are congruent, the quadrilateral is a parallelogram by Theorem 6-6 . y 80 9 y 89 89 , then ABCD is a parallelogram. Geometry Lesson 6-3 Daily Notetaking Guide DailyNotetaking Guide Geometry Lesson 6-3 115 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Quick Check.Lesson 6-4 1. Find the values of a and c for which PQRS must be a parallelogram. Q R (a 40) a 3c 3 a 70, c 2 c1 P Special Parallelograms Lesson Objectives 1 Use properties of diagonals of rhombuses and rectangles 2 Determine whether a parallelogram is a rhombus or a rectangle NAEP 2005 Strand:Geometry Topic: Geometry Local Standards: ____________________________________ S 2. Can you prove the quadrilateral is a parallelogram? Explain. a. Given: PQ > SR, PQ 6SR Prove: PQRS is a parallelogram. Q Yes; a pair of opposite sides are parallel and congruent (Theorem 6-8). RRhombuses Theorem 6-9 B Each diagonal of a rhombus bisects two angles of the rhombus. AC bisects BAD, so 1 2 AC bisects BCD , so 3 4 x AC S b. Given: DH > GH, EH > FH Prove: DEFG is a parallelogram. D E H G F © Pearson Education, Inc., publishing as Pearson Prentice Hall. P The diagonalsof a rhombus are ' A perpendicular D A congruent C . Rectangles Theorem 6-11 AC C 4 B BD The diagonals of a rectangle are 3 1 2 Theorem 6-10 x No; the diagonals do not necessarily bisect each other—the figure could be a trapezoid, with DG and EF being parallel but of unequal length. All rightsreserved. Key Concepts. All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. A © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Examples. for which ABCD must be a parallelogram. D Theorem 6-6 states If both pairs of opposite angles ofa quadrilateral are , then the quadrilateral is a parallelogram. 1 Finding Values for Parallelograms Find values for x and y A ( congruent C b. Given: mE mG 75°, mF 105° Prove: EFGH is a parallelogram. If x represents the measure of the unmarked angle, 360 105 x 75 105 75 , so x = . © PearsonEducation, Inc., publishing as Pearson Prentice Hall. opposite angles then the quadrilateral is a parallelogram. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Theorem 6-6 If both pairs of All rights reserved. All rights reserved. Theorem 6-5 If both pairs of opposite sides of a quadrilateralare congruent, then B D A D B C . BD Parallelograms Theorem 6-12 If one diagonal of a parallelogram bisects two angles of the parallelogram, then the parallelogram is a rhombus. Theorem 6-13 If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. Theorem 6-14 Ifthe diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. 116 Geometry Lesson 6-3 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 6-4 Geometry 117 29 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Examples. Example. 1 Finding Angle Measures Find themeasures of the 3 Identifying Special Parallelograms The diagonals of ABCD are such that numbered angles in the rhombus. Theorem 6-9 states that each diagonal of a rhombus bisects two angles of the rhombus, so m1 78. B 78 the diagonals of the rhombus are perpendicular so m2 90 2 3 Theorem 6-10 states that , A AC 16 cm and BD 8 cm. Can you conclude that ABCD is a rhombus or a rectangle? Explain. C 1 ABCD 4 D be a rectangle, because AC BD and the diagonals of a cannot rectangle are (Theorem 6-11). ABCD may be a rhombus, congruent AC ' BD but it may also not be one, dependingon whether (Theorem 6-10). . m3 90 78 . Because mABD 78, complementary All rights reserved. 3 and ABD must be . 12 Finally, because BC DC, the Isosceles Triangle Theorem allows you to conclude that 1 4. So m4 . 78 All rights reserved. Because the four angles formed by the diagonals all musthave measure 90, Quick Check. 2. Find the length of the diagonals of rectangle GFED if FD 5y 9 and GE = y 5. F E G D 81 2 2 Finding Diagonal Length One diagonal of a rectangle has length 8x 2. The other diagonal has length 5x 11. Find the length of each diagonal. congruent By Theorem 6-11, thediagonals of a rectangle are 8x 2 Subtract 3 x ( )2 5x 11 5( 3 ) 11 8x 2 8 3 5x from each side. Subtract 2 from each side. Divide each side by 3 . 26 Substitute. 26 The length of each diagonal is Substitute. 26 . Quick Check. 1. Find the measures of the numbered angles in the rhombus. 50 1 m⬔1 90, m⬔250, m⬔3 50, m⬔4 40 4 118 Geometry Lesson 6-4 2 3 3. A parallelogram has angles of 30°, 150°, 30°, and 150°. Can you conclude that it is a rhombus or a rectangle? Explain. No; there is not enough information to conclude that the parallelogram is a rhombus. It cannot be a rectangle because it has noright angles. Daily Notetaking Guide Daily Notetaking Guide All rights reserved. 2 Geometry Lesson 6-4 119 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date ________________ Lesson 6-5 Examples. 1 Finding Angle Measures in Trapezoids WXYZ is Trapezoids and Kites Lesson Objective 1 Verify and use properties of trapezoids and kites NAEP 2005 Strand: Geometry Topic: Relationships Among GeometricFigures Local Standards: ____________________________________ are 156 Trapezoids Theorem 6-15 diagonals of an isosceles trapezoid are congruent. All rights reserved. All rights reserved. Theorem 6-16 The mW 180 mW . Z a leg of a trapezoid 180 Vocabulary and Key Concepts. congruent Y WTwo angles that share mX mW The base angles of an isosceles trapezoid are X 156 an isosceles trapezoid, and mX 156. Find mY, mZ, and mW. . supplementary Substitute. 24 Subtract 156 from each side. congruent Because the base angles of an isosceles trapezoid are mY mX 156 and mZ m⬔W , 24. A 2 Using Isosceles Trapezoids Half of a spider’s web is shown at the right, C formed by layers of congruent isosceles trapezoids. Find the measures of the angles in ABDC. The base angles of a trapezoid are two angles that share a base of the trapezoid. B D Trapezoid ABDC is part of an isoscelestriangle whose vertex is at angles Base angles Leg Kites Theorem 6-17 The diagonals of a kite are perpendicular . © Pearson Education, Inc., publishing as Pearson Prentice Hall. Leg Base © Pearson Education, Inc., publishing as Pearson Prentice Hall. the center of the web. Because there are 6adjacent congruent vertex angles at the center of the web, together forming a straight angle, each vertex angle measures 180 , or 30 . 6 180 By the Triangle Angle-Sum Theorem, mA mB 30 so mA mB 150 , . Because ABDC is part of an isosceles triangle, mA mB, so 2(mA) 150 and mA mB 75 . Anotherway to find the measure of each acute angle is to divide the difference between 180 and the measure of the vertex angle by 2: 180 2 30 75 . 2 Because the bases of a trapezoid are parallel, the two angles that share a leg are supplementary , so mC mD 180 75 = 105 . Quick Check. 1. In the isoscelestrapezoid, mS 70. Find mP, mQ, and mR. P Q 110, 110, 70 S 120 30 Geometry Lesson 6-5 Geometry Daily Notetaking Guide Daily Notetaking Guide 70 Geometry Lesson 6-5 R 121 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3x 3x © Pearson Education,Inc., publishing as Pearson Prentice Hall. 11 9 . Diagonals of a rectangle are congruent. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 5x 11 Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date________________ Example. Lesson 6-6 3 Finding Angle Measures in Kites Find m1, m2, and m3 in the kite. Lesson Objective S 1 1 3Example. 1 Proving Congruency Show that TWVU is a parallelogram by U RU RS Definition of a kite m1 72 Isosceles Triangle Theorem m3 mRDU 72 180 mRDU 90 perpendicular 90 72 180 162 180 m3 18 Triangle Angle-Sum Theorem Diagonals of a kite are perpendicular. If both pairs of opposite sidesof a quadrilateral are congruent from each side. Quick Check. ( ( WV ( TU ( 3. Find m1, m2, and m3 in the kite. m⬔1 90, m⬔2 46, and m⬔3 44 46 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Themeasure of both inner angles is 85. The measure of both outer angles is 95. ( E TU . Use the distance formula. ) ( ) ( ) a , b , W a c , b d , V c e , d , and U ) ( ) )2 ( d 0 )2 c 2 d 2 2 2 ac ce ) ( bd d ) 2 2 a e ) ( b 0 ) (a e) 2 b 2 VU 2. The middle ring of the piece of ceiling shown is made from congruentisosceles trapezoids. Imagine a circular glass ceiling made from the ceiling pieces with 18 angles meeting at the center. What are the measures of the two sets of base angles of the trapezoids in the middle ring? x U(e, 0) O You can prove that TWVU is a parallelogram by showing that TW V(c e, d) .Theorem 6-5 Use the coordinates T Simplify. 162 W(a c, b d) C T(a, b) , then the quadrilateral is a parallelogram by TW VU and WV Substitute. Subtract A y proving pairs of opposite sides congruent. . All rights reserved. Diagonals of a kite are All rights reserved. m2 90 m3 Topic: Position and Direction T2 72 m3 NAEP 2005 Strand: Geometry Name coordinates of special figures by using their properties Local Standards: ____________________________________ D R Placing Figures in the Coordinate Plane ac a ce e 2 2 b bd Because TW VU and WV TU, TWVU is a c2 ( ) e , 0 . d2 parallelogram (a e)2 b 2 . Quick Check. 1. Use the diagram above. Use a different method: Show that TWVU is a parallelogram by finding the midpoints of the diagonals. Midpoint of TV 5 aa 1 c 1 e , b 1 d b midpoint of UW . Thus, the 2 2 diagonals bisect each other, and TWVU is a parallelogram. 1 3 122 Geometry Lesson6-5 Geometry Lesson 6-6 123 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Example. Lesson 6-7 2 NamingCoordinates Use the properties of parallelogram y OCBA to find the missing coordinates. Do not use any new variables. (0, 0) The vertex O is the origin with coordinates O . B Proofs Using Coordinate Geometry Lesson Objective 1 Prove theorems using figures in the coordinate plane (p, q) A NAEP 2005Strand: Geometry Topics: Position and Direction; Mathematical Reasoning Local Standards: ____________________________________ Because point A is p units to the left of point O, point B Vocabulary and Key Concepts. is also p units to the left of point C , because OCBA is a p x . Because AB 6 COand CO is horizontal, AB is C (x, 0) O . So point B has the same second coordinate, q , as point A. The missing coordinates are O (0, 0) and B (p x, q) Theorem 6-18: Trapezoid Midsegment Theorem All rights reserved. also horizontal x . All rights reserved. parallelogram. So the first coordinate of point Bis parallel (1) The midsegment of a trapezoid is Quick Check. lengths of the bases . opposite sides of the trapezoid. 2. Use the properties of parallelogram OPQR to find the missing coordinates. Do not use any new variables. y R(b, c) Q(?, ?) R Q(s b, c) M T P(s, 0) Geometry Lesson 6-6 All-In-OneAnswers Version A A N P x Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. O 124 to the bases. (2) The length of the midsegment of a trapezoid is half the sum of the The midsegment of a trapezoid is the segment that joins the midpoints of the nonparallel ©Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Daily Notetaking Guide Daily Notetaking Guide MN 6 TP , MN 6 1 RA , and MN 2 ( TP RA ) Examples. 1 Planning a Coordinate Geometry Proof Examine trapezoid y R(2b, 2c)A(2d, 2c) TRAP. Explain why you can assign the same y-coordinate to points R and A. M In a trapezoid, only one pair of sides is parallel. In TRAP, TP 6 must be RA . Because TP lies on the horizontal x-axis, RA also horizontal T N x P(2a, 0) . The y-coordinates of all points on a horizontal line are thesame, so points R and A have the same y-coordinates. Daily Notetaking Guide Geometry Lesson 6-7 Geometry 125 31 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________Name_____________________________________ Class____________________________ Date ________________ 2 Using Coordinate Geometry Use coordinate geometry to prove that the Quick Check. quadrilateral formed by connecting the midpoints of rhombus ABCD is a rectangle. Drawquadrilateral XYZW by connecting the midpoints of ABCD. Given: MN is the midsegment of trapezoid TRAP. 1 Prove: MN 6 TP, MN 6 RA, and MN = 2 (TP + RA) y M a. Find the coordinates of midpoints M and N. How do the multiples T of 2 help? D(0, 2b) Z(a, b) y R(2b, 2c) A(2d, 2c) 1. Complete theproof of Theorem 6-18. W(a, b) C(2a, 0) N x P(2a, 0) M(b, c) and N(a d, c); starting with multiples of 2 eliminates fractions when using the midpoint formula. A(2a, 0) All rights reserved. B(0, 2b) From Lesson 6-6, you know that XYZW is a parallelogram. congruent If the diagonals of a parallelogram are Allrights reserved. x X(a, b) Y(a, b) b. Find and compare the slopes of MN, TP, and RA. 0, 0, 0; they are all parallel. , then the parallelogram is a rectangle from Theorem 6–14. To show that XYZW is a rectangle, find the lengths of its diagonals, and then equal compare them to show that they are (a a)2 (b b)2( 2a )2 ( 2b )2 4a2 4b2 YW XZ YW Because the diagonals are parallelogram XYZW is a rectangle 126 congruent, . Geometry Lesson 6-7 d. In parts (b) and (c), how does placing a base along the x-axis help? The base along the x-axis allows calculating length by subtracting x-values. 2. Use the diagramfrom Example 2. Explain why the proof using A(2a, 0), B(0, 2b), C(2a, 0), and D(0, 2b) is easier than the proof using A(a, 0), B(0, b), C(a, 0), and D(0, b). Using multiples of 2 in the coordinates for A, B, C, and D eliminates the use of fractions when finding midpoints, since finding midpoints requires divisionby 2. Daily Notetaking Guide Daily Notetaking Guide All rights reserved. MN d a b, TP 2a, RA 2d 2b. So RA TP 2d 2a 2b which is twice MN. So the midsegment is half the sum of the bases. (a a) (b (b)) ( 2a )2 ( 2b )2 4a2 4b2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. c. Find andcompare the lengths MN, TP, and RA. . 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. XZ 2 Geometry Lesson 6-7 127 Name_____________________________________ Class____________________________ Date________________Name_____________________________________ Class____________________________ Date ________________ Lesson 7-1 2 Properties of Proportions b Complete: If a4 5 12 b , then 12 5 NAEP 2005 Strand: Geometry Topic: Position and Direction ab 5 Local Standards:____________________________________ ab (3) ac b d b (4) a 1 b c1d d A proportion is a statement that two ratios are equal. All rights reserved. d c (2) ba All rights reserved. c a b d is equivalent to bc c a b d and a : b c : d are examples of proportions. 6 1 4 24 16 4 is an example of an extendedproportion. a. 25 n 5 35 5n 5 2 ( ) 35 Simplify. n5 14 Divide each side by 5 . 3x 5 A scale drawing is a drawing in which all lengths are proportional to corresponding actual lengths. A scale is the ratio of any length in a scale drawing to the corresponding actual length. The lengths may be in different units.Examples. 1 Finding Ratios A scale model of a car is 4 in. long. The actual car is 15 ft long. What is the ratio of the length of the model to the length of the car? Write both measurements in the same units. 180 15 ft 15 12 in. in. length of model 4 in. 4 in. 4 length of car 5 15 ft 5 180 in. 5 180 5 ( 2x x 1 1Geometry ) 1 2 Subtract Property Cross-Product Property Distributive Property 2x from each side. Quick Check. 1. A photo that is 8 in. wide and 531 in. high is enlarged to a poster that is 2 ft wide and 113 ft high. What is the ratio of the height of the photo to the height of the poster? 1:3 n 2. Write twoexpressions that are equivalent to m 4 5 11. 4 5 11, m 1 4 5 n 1 11 Answers may vary. Sample: m n 4 11 1 45 The ratio of the length of the scale model to the length of the car is Geometry Lesson 7-1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. a d b c © Pearson Education, Inc.,publishing as Pearson Prentice Hall. a c b d Cross-Product 70 x5 2 extremes 32 a 5n 5 to the product of the means. means . 3 Solving for a Variable Solve each proportion. The Cross-Product Property states that the product of the extremes of a proportion is equal a:b c:d 12a Divide each side by Simplify.1 x b. x 1 3 52 3x 5 2 An extended proportion is a statement that three or more ratios are equal. 128 12a 4 b 12 5 Properties of Proportions . Cross-Product Property 48 12a Vocabulary and Key Concepts. (1) ad 48 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Ratios and ProportionsLesson Objective 1 Write ratios and solve proportions 1 : 45 . Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 7-1 129 All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Example. Lesson 7-2 4 Using Proportions Two cities are 3 12 in. apart on a map with the scale 1 in. 50 mi. Find theactual distance. Let d represent the actual distance. map distance (in.) actual distance (mi.) 5 Similar figures have the same shape but not necessarily the same size. Two polygons are ) 31 2 Cross-Product 175 Property Simplify. 175 The cities are actually miles apart. All rights reserved. d5 ( Quick Check.The mathematical symbol for similarity is ⬃ . The similarity ratio is the ratio of the lengths of corresponding sides of similar figures. similar to the original rectangle. 6 b. n 18 165n z 0.75 The golden ratio is the ratio of the length to the width of any golden rectangle, about 1.618 : 1. 4. Recall Example 4. Youwant to make a new map with a scale of 1 in. 35 mi. Two cities that are actually 175 miles apart are to be represented on your map. What would be the distance in inches between the cities on the new map? 5 inches © Pearson Education, Inc., publishing as Pearson Prentice Hall. n3 © Pearson Education,Inc., publishing as Pearson Prentice Hall. All rights reserved. similar if corresponding angles are congruent and corresponding sides are proportional. A golden rectangle is a rectangle that can be divided into a square and a rectangle that is 3. Solve each proportion. a. 5z 5 20 3 Example. 1 UnderstandingSimilarity 䉭ABC 䉭XYZ. Y ⬔B ⬔ Y and m⬔Y 78, so m⬔B 78 because congruent angles have the same measure. A C X BC 5 b. YZ XZ AC 78 42 Z BC AC . corresponds to XZ, so YZ XZ Quick Check. 1. Refer to the diagram for Example 1. Complete: BC and YZ 42 AB XY Daily Notetaking Guide DailyNotetaking Guide Geometry Lesson 7-2 131 Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date ________________ Examples. QuickCheck. 2 Determining Similarity Determine whether the K parallelograms are similar. Explain. Check that corresponding sides are proportional. B 1 2 1 2 DA MJ N A Corresponding sides of the two parallelograms proportional. are 2 4 4 Y 21 D 1 14 J M 2 Check that corresponding angles are congruent.⬔B corresponds to ⬔ K , but m⬔B ⫽ m⬔K, so corresponding are not angles congruent. Although corresponding sides parallelograms are not are proportional, the Because 䉭ABC 䉭YXZ, you can write and solve a proportion. x Corresponding sides are XZ proportional 12 x . 12 C X 40 Z 50 Solve for x.15 Simplify. 4 Using the Golden Ratio The dimensions of a rectangular tabletop are in the Golden Ratio. The shorter side is 40 in. Find the longer side. Let ᐍ represent the longer side of the tabletop. 1.618 ᐉ 40 5 1 64.72 ᐉ5 Write a proportion using the The table is about 132 Cross-Product 65 24 M 50Substitute. 40 x5 B 30 x 12 40 P Y A BC X 9 3 Using Similar Figures If 䉭ABC 䉭YXZ, find the value of x. 50 16 Yes; corresponding angles are congruent and corresponding sides are proportional. . the corresponding angles are not congruent AC 5 YZ Z 18 12 3. Refer to the diagram for Example 3. FindAB. similar because Golden Ratio © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2 4 C 120 2 CD LM 100 All rights reserved. BC KL 2. Sketch 䉭XYZ and 䉭MNP with ⬔X ⬔M, ⬔Y ⬔N, and ⬔Z ⬔P. Also, XY 12, YZ 14, ZX 16, MN 18, NP 21, and PM 24. Can you conclude that the twotriangles are similar? L All rights reserved. 2 4 AB 5 JK 1 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Geometry Lesson 7-1 B Complete each statement. a. m⬔B m⬔A 130 Topics: Transformation of Shapes andPreservation of Properties; Measuring Physical Attributes Vocabulary. Substitute. 50 d 5 50 Identify similar polygons Apply similar polygons Local Standards: ____________________________________ 1 d 1 2 NAEP 2005 Strand: Geometry and Measurement 1 in. 50 mi All rights reserved. 31 2 SimilarPolygons Lesson Objectives 4. A golden rectangle has shorter sides of length 20 cm. Find the length of the longer sides. about 32.4 cm . Property in. long. Geometry Lesson 7-2 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 7-2 Geometry 133 33Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 7-3 2 Using Similarity Theorems Explain why the triangles Proving Triangles Similar W must be similar. Write a similarity statement. Lesson Objectives Use AA, SAS, and SSS similarity statements Apply AA, SAS, and SSS similarity statements Topic: Transformation of Shapes and Preservation ofProperties 1 VY 12 VW 5 24 Local Standards: ____________________________________ VZ 5 and VX 2 18 36 Z 18 1 V , 2 12 Y so corresponding sides are proportional. 36 X Vocabulary and Key Concepts. R PLM P T S L M All rights reserved. 䉭TRS 䉭 Therefore, 䉭YVZ 䉭WVX Side-Angle-SideSimilarity (SAS) by the All rights reserved. Postulate 7-1: Angle-Angle Similarity (AA⬃) Postulate If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. Theorem. 3 Using Similar Triangles Joan places a mirror 24 ft from the base of a T tree. When shestands 3 ft from the mirror, she can see the top of the tree reflected in it. If her eyes are 5 ft above the ground, how tall is the tree? TR represents the height of the tree, Point M represents the mirror, and point J represents Joan’s eyes. Both Joan and the tree are Theorem 7-1: Side-Angle-Side Similarity(SAS⬃) Theorem If an angle of one triangle is congruent to an angle of a second triangle, and the sides including the two angles are proportional, then so m⬔JOM m⬔TRM, and therefore . lJOM O lTRM The light reflects off the mirror at the same angle at which it hits the mirror, so ⬔JMO ⬔ TMR . thetriangles are similar. Indirect measurement is a way of measuring things that are difficult to measure directly. Examples. 1 Using the AA⬃ Postulate MX ' AB. Explain why the triangles M are similar. Write a similarity statement. K . ⬔A > ⬔ B because their measures are equal. 䉭AMX 䉭BKX Angle-AngleSimilarity (AA) by the 134 Postulate. © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. the triangles are similar. BXK J 5 O 3 M 24 R Geometry Lesson 7-3 135 Use similar triangles to find the height of the tree. Theorem 7-2:Side-Side-Side Similarity (SSS⬃) Theorem If the corresponding sides of two triangles are proportional, then Because MX ' AB, ⬔AXM and ⬔BXK are right angles , so ⬔AXM > ⬔ to the ground, perpendicular nJOM , nTRM TR AA Postulate RM TR JO Corresponding sides of similar triangles areproportional. OM All rights reserved. 2 24 ⬔YVZ > ⬔WVX because they are vertical angles. 24 Substitute. 3 5 24 TR 5 Solve for TR. 3 TR 5 40 58 A Geometry Lesson 7-3 40 The tree is 58 feet tall. B X Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Quick Check. Lesson 7-4 1. In Example 1, you have enough information to write a similarity statement. Do you haveenough information to find the similarity ratio? Explain. Similarity in Right Triangles Lesson Objective 1 Find and use relationships in similar right triangles No; none of the side lengths are given. NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties LocalStandards: ____________________________________ 2. Explain why the triangles at the right must be similar. Write a similarity statement. G 8 F 9 A 6 B 6 All rights reserved. Theorem 7-3 12 E All rights reserved. nABC , nEFG Vocabulary and Key Concepts. 8 AC 5 CB 5 AB 5 3 , so the triangles aresimilar by the SSS~ Theorem; EF 4 EG GF C The altitude to the hypotenuse of a right triangle divides the triangle into similar two triangles that are to the original triangle and to each other. similar Corollary 1 to Theorem 7-3 The length of the altitude to the hypotenuse of a right triangle is the geometricmean of the lengths of the segments of the hypotenuse. 3. In sunlight, a cactus casts a 9-ft shadow. At the same time, a person 6 ft tall casts a 4-ft shadow. Use similar triangles to find the height of the cactus. x ft 6 ft 4 ft 9 ft 13.5 ft The altitude to the hypotenuse of a right triangle separates the hypotenuse© Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Corollary 2 to Theorem 7-3 so that the length of each leg of the triangle is the geometric mean of the adjacent hypotenuse segment length of the and the length of thehypotenuse . The geometric mean of two positive numbers a and b is the positive number x such a 5 x. that x b Examples. 1 Finding the Geometric Mean Find the geometric mean of 3 and 12. x 3 x5 x2 5 Write a proportion. 12 36 x5 x5 6 !36 Cross-Product Property. Find the positive square root. Thegeometric mean of 3 and 12 is 6 . 136 34 Geometry Lesson 7-3 Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 7-4 137 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 NAEP 2005 Strand: Geometry Geometry: All-In-OneAnswers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Cup 2 Finding Distance At a golf course, Maria drove her ball 192 yd straight y toward the cup. Her brother Gabriel drove his ball straight 240 yd, but nottoward the cup. The diagram shows the results. Find x and y, their remaining distances from the cup. Use Corollary 2 of Theorem 7-3 to solve for x. 240 Cross-Product Property y Distributive Property If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sidesproportionally. Theorem 7-5: Triangle-Angle-Bisector Theorem If a ray bisects an angle of a triangle, then it divides the opposite side into two segments that y Substitute 108 are proportional to the other two sides of the triangle. for x. 108 300 y y Corollary to Theorem 7-4 Simplify. y Cross-Product !32,400Find the positive square root. Property y 180 Maria’s ball is yd from the cup, and Gabriel’s ball is 108 yd 180 from the cup. Quick Check. 1. Find the geometric mean of 15 and 20. 2. Recall Example 2. Find the distance between Maria’s ball and Gabriel’s ball. 10"3 144 yd © Pearson Education, Inc.,publishing as Pearson Prentice Hall. 32,400 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 108 y2 If three parallel lines intersect two transversals, then the segments intercepted on the transversals are proportional. Name_____________________________________Class____________________________ Date ________________ R 6 x 9 15 y5 x5 15 A 10 6 Substitute. 6 y 72 y 7.2 Cross-Product Property Solve for y. Geometry Lesson 7-5 Example. 139 G 40 24 I 12.5 x N NA Corollary to the Side-Splitter Theorem 75 15y C 10 Theorem 3 Using the Triangle-Angle-Bisector Theorem Find the value of x. Cross-Product x IG GH Solve for x. 12.5 Corollary to the Side-Splitter Theorem y 12.5 Substitute y for x. Cross-Products 112.5 7.5 and y 5 15 24 IK Triangle-Angle-Bisector Theorem. HK x Substitute. 30 ( 30 ) x 40 Property Solve for y. 40 24 H 30 K Property All rightsreserved. 15 Side-Splitter 12.5 y T 9 U 5 x x 9 . 12 M CN Name_____________________________________ Class____________________________ Date ________________ All rights reserved. 6 x5 UT y B CM MB S 5 x Solve for x. 18 Quick Check. 7.5 3. Find the value of y. 3.6 5.76 1. Use the Side-Splitter Theorem to find the value of x. 5 x 1.5 2.5 x 2. Solve for x and y. x ⫽ 225 13 ; y ⫽ 28.6 16.5 15 y 26 x © Pearson Education, Inc., publishing as Pearson Prentice Hall. Quick Check. 1.5 y 5 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing asPearson Prentice Hall. Because RSTU is a trapezoid, RS 6 c d Examples. Daily Notetaking Guide Daily Notetaking Guide 2 Using the Corollary to the Side-Splitter Theorem The segments joining the sides of trapezoid RSTU are parallel to its bases. Find x and y. a b5 d 1 Using the Side-Splitter TheoremFind y. 12 y Geometry Lesson 7-4 c a b 10 5 Local Standards: ____________________________________ ★ Solve for x. Write a proportion. y 138 Topic: Transformation of Shapes and Preservation of Properties Theorem 7-4: Side-Splitter Theorem 20,736 x 108 192 All rights reserved. Use the Side-Splitter Theorem Use the Triangle-Angle-Bisector Theorem Key Concepts. Use Corollary 2 of Theorem 7-3 to solve for y. y NAEP 2005 Strand: Geometry 192 x 5 108 x 192 1 All rights reserved. 192 x 240 Proportions in Triangles Lesson Objectives 192 57,600 36,864 Lesson 7-5 x M 2 Write a proportion.192 (x 1 192) 5 2402 192x G All rights reserved. x 192 240 Name_____________________________________ Class____________________________ Date________________ 8 30 140 Geometry Lesson 7-5 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide GeometryLesson 7-5 Geometry 141 35 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 8-1 1 Example. The Pythagorean Theorem and Its Converse Lesson Objectives 1Pythagorean Triples Find the length of the hypotenuse of #ABC. NAEP 2005 Strand: Geometry Use the Pythagorean Theorem Use the Converse of the Pythagorean Theorem 2 Name_____________________________________ Class____________________________ Date ________________ a2 1b2 5 c2 Local Standards: ____________________________________ Î 169 a . b a2 b2 c 2 Theorem 8-2: Converse of the Pythagorean Theorem All rights reserved. All rights reserved. hypotenuse is equal to the square of the length of the c If the square of the length of one side of a triangle is equal to thesum of c Take the square root. A 5 C Simplify. The length of the hypotenuse is 13 . The lengths of the sides, 5, 12, and 13 , form a Pythagorean triple because they are whole numbers that satisfy a2 1 b2 5 c2. Quick Check. the squares of the lengths of the other two sides, then the triangle is a right 12Simplify. 13 c legs Theorem. Substitute 5 for a, and 12 for b. 169 c2 Theorem 8-1: Pythagorean Theorem Pythagorean Use the 52 122 c2 Vocabulary and Key Concepts. In a right triangle, the sum of the squares of the lengths of the B Do the lengths of the sides of #ABC form a Pythagorean triple? Topic:Relationships Among Geometric Figures 1. A right triangle has a hypotenuse of length 25 and a leg of length 10. Find the length of the other leg. Do the lengths of the sides form a Pythagorean triple? triangle. 5!21; no obtuse is c a . C If c 2 a 2 b 2, the triangle is obtuse A b . Theorem 8-4 B If the squareof the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, the triangle acute is a c C b . If c 2 a 2 b 2, the triangle is acute A . All rights reserved. B the sum of the squares of the lengths of the other two sides, the triangle © Pearson Education, Inc.,publishing as Pearson Prentice Hall. If the square of the length of the longest side of a triangle is greater than © Pearson Education, Inc., publishing as Pearson Prentice Hall. Theorem 8-3 A Pythagorean triple is a set of nonzero whole numbers a, b, and c that satisfy the equation a 2 ⫹ b 2 ⫽ c 2.Geometry Lesson 8-1 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ Examples. 3 Is It a Right Triangle? Is this triangle a right triangle? answer in simplest radical form 12 Use the 82b2 122 Substitute 8 for a, 12 for c. 64 b2 144 Simplify. b2 80 b 42 62 0 72 Substitute 4 for a, 6 for b, and 7 for c. 16 36 0 49 Simplify. 52 2 49 Subract 64 from both sides. Because a2 b2 ⫽ c2, the triangle Take the square root. ( 5 ) Simplify. b 4 !5 c2 0 a2 1 b2 Geometry Daily Notetaking Guide © PearsonEducation, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 36 Compare c 2 with a 2 b 2. 202 0 10 2 15 2 Substitute the greatest length for c . 400 0 100 225 Simplify. 400 . 325 Because c2 ⬎ a2 b2, the triangle is a(n) Geometry Lesson 8-1 a righttriangle. and 20 represent the lengths of the sides of a triangle. Classify the triangle as acute, obtuse, or right. 2. The hypotenuse of a right triangle has length 12. One leg has length 6. Find the length of the other leg. Leave your answer in simplest radical form. 6 !3 is not 7m 6m 4 Classifying Triangles asAcute, Obtuse, or Right The numbers 10, 15, Quick Check. 144 4m a2 1 b2 0 c2 All rights reserved. b Î 16 8 Theorem. All rights reserved. b Î 80 Pythagorean 143 Name_____________________________________ Class____________________________ Date ________________ Example. 2 UsingSimplest Radical Form Find the value of b. Leave your a2 1 b2 5 c2 Geometry Lesson 8-1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 142 obtuse triangle. Quick Check. 3. A triangle has sides of lengths 16, 48, and 50. Is the triangle a right triangle? no 4. A triangle has sides of lengths7, 8, and 9. Classify the triangle by its angles. acute Daily Notetaking Guide Geometry Lesson 8-1 145 All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 8-2 2 Applying the 45-45-90 Triangle Theorem The distance from Special Right Triangles Lesson Objectives NAEP 2005 Strand: Geometry Use theproperties of 45-45-90 triangles Use the properties of 30-60-90 triangles 1 2 Topic: Relationships Among Geometric Figures Local Standards: ____________________________________ and the hypotenuse "2 ? 45 s s2 length of the hypotenuse is "2 times the length of a leg. 45 s leg All rights reserved.congruent In a 45-45-90 triangle, both legs are All rights reserved. Theorem 8-5: 45-45-90 Triangle Theorem hypotenuse 2 ? 60 s shorter leg 1 Finding the Length of the Hypotenuse Find the value of the variable. Use the 45-45-90 Triangle Theorem to find the hypotenuse. hypotenuse "2 "12 45 h 5 6 ? legSimplify. 45 4(3) h 5( 2 ) 3 h5 5 6 "3 The length of the hypotenuse is 10"3 Divide each side by "2 . 67.882251 Use a calculator. Each side of the playground is about 68 ft. 3 Using the Length of One Side Find the value of each variable. Use the 30-60-90 Triangle Theorem to find the lengths of the legs.hypotenuse 2 ? shorter leg 4"3 "3 y Simplify. 60 y "3 ? shorter leg y "3 ? 2 !3 30°-60°-90° Substitute y 2 ? "3 ? "3 y 30 4 3 Divide each side by 2 . 2 x 2 Examples. h "2 ? 5!6 leg 30 s3 2s . shorter leg longer leg "3 ? Triangle Theorem 2"3 x for shorter leg. Simplify. 6 2"3 The length of the shorter leg is 6 ,and the length of the longer leg is . Quick Check. 1. Find the length of the hypotenuse of a 45-45-90 triangle with legs of length 5 !3. 5!6 2. A square garden has sides of 100 ft long. You want to build a brick path along a diagonal of the square. How long will the path be? Round your answer to the nearestfoot. 141 ft . Geometry Lesson 8-2 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 8-2 147 Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________Class____________________________ Date________________ Example. 4 Applying the 30-60-90 Triangle Theorem A rhombus-shaped Lesson 8-3 25 garden has a perimeter of 100 ft and a 60 angle. Find the area of the garden to the nearest square foot. Because a rhombus has four sides of equallength, each side is 25 ft h 25 ft. Local Standards: ____________________________________ Vocabulary. The tangent of acute ⬔A in a right triangle is the ratio of the length of the leg opposite The height h is the longer leg of the right triangle. To find the height h, you can use the properties of 30-60-90triangles. Then apply the area formula. Divide each side by 2 . h 12.5 "3 longer leg "3 ? shorter leg A bh Use the formula for the area of a A A ( 25 )(12.5!3) 541.26588 To the nearest square foot, the area is Substitute 25 for b and parallelogram 12.5"3 A Leg ⬔A length of leg opposite lA length of legadjacent to lA tangent of ⬔A opposite adjacent ft 2. 8 x 4, y 4 !3 30 60 x y 4. A rhombus has 10-in. sides, two of which meet to form the indicated angle. Find the area of each rhombus. (Hint: Use a special right triangle to find height.) a. a 30 angle b. a 60 angle Geometry Lesson 8-2 All-In-One AnswersVersion A Daily Notetaking Guide . Examples. 1 Writing Tangent Ratios Write the tangent ratios for ⬔A and ⬔B. B 20 C tan A tan B 50 !3 in.2 ⬔A opposite C adjacent to You can abbreviate the equation as tan A 3. Find the value of each variable. 148 Leg for h. Use a calculator. 541 B . Quick Check. 50in.2 All rights reserved. 12.5 hypotenuse 2 ? shorter leg © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2 lA to the length of the leg adjacent to lA. All rights reserved. 25 shorter leg NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes 60° Because a rhombus is also aparallelogram, you can use the area formula A bh . Draw the altitude h, and then solve for h. 25 2 ? shorter leg The Tangent Ratio Lesson Objective 1 Use tangent ratios to determine side lengths in triangles © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. shorter leg ©Pearson Education, Inc., publishing as Pearson Prentice Hall. . The length of the longer leg is "3 times the length of the © Pearson Education, Inc., publishing as Pearson Prentice Hall. 96 96 ft "2 ? leg hypotenuse "2 x © Pearson Education, Inc., publishing as Pearson Prentice Hall. shorter leg the lengthof the 146 s twice In a 30-60-90 triangle, the length of the hypotenuse is 10 96 "2 ? leg 4 !3 2 ? x Theorem 8-6: 30-60-90 Triangle Theorem h s leg Key Concepts. h 5 one corner to the opposite corner of a square playground is 96 ft. To the nearest foot, how long is each side of the playground? Thedistance from one corner to the opposite corner, 96 ft, is the length of the hypotenuse of a 45-45-90 triangle. 29 21 A opposite adjacent opposite adjacent Daily Notetaking Guide BC AC AC BC 20 21 21 20 Geometry Lesson 8-3 Geometry 149 37 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date ________________ 2 Using a Tangent Ratio To measure the height of a tree, Almawalked Quick Check. 125 ft from the tree and measured a 32 angle from the ground to the top of the tree. Estimate the height of the tree. 1. a. Write the tangent ratios for ⬔K and ⬔J. J 3, 7 7 3 3 L right The tree forms a K 7 32 125 ft angle with the ground, so you can use the Use the tangent ratio. 125 (height 125 ) tan 32° 78.108669 32 78 The tree is about Solve for height. All rights reserved. height tan 32 125 All rights reserved. tangent ratio to estimate the height of the tree. b. How is tan K related to tan J? They are reciprocals. Use a calculator. 2. Find the value of w to the nearest tenth. a. b. 10 54feet tall. 3 Using the Inverse of Tangent Find m⬔R to the nearest degree. c. w 1.0 w w 2.5 28 57 33 S T 47 tan R 47 41 1 m⬔R tan 47 ⫺1 ( ) 47 41 1.14634146 48.900494 41 1.14634146 1.9 13.8 3.8 3. Find m⬔Y to the nearest degree. 100 P T 68 41 Y All rights reserved. 41 © Pearson Education, Inc.,publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. R So m⬔R 49 . Geometry Lesson 8-3 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________Date________________ Lesson 8-4 Name_____________________________________ Class____________________________ Date ________________ 2 Using the Cosine Ratio A 20-ft wire supporting a flagpole forms a Sine and Cosine Ratios Lesson Objective 1 Use sine and cosine to determineside lengths in triangles 151 Geometry Lesson 8-3 35 angle with the flagpole. To the nearest foot, how high is the flagpole? NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes The flagpole, wire, and ground form a wire as the Local Standards:____________________________________ with the right triangle 20 ft 35 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 150 . Because you know an angle and the hypotenuse measure of its hypotenuse, you can use the ratio cosine to find the height of the flagpole. Vocabulary. cosineof ⬔A B leg opposite lA hypotenuse This can be abbreviated sin A hypotenuse opposite hypotenuse Leg ⬔A C . A Leg adjacent hypotenuse . A T sin T cos T 38 adjacent hypotenuse 20 5 16 4 5 20 16 20 5 adjacent cos G hypotenuse 12 3 20 Geometry Lesson 8-4 Geometry 16 R 20 12 G 4 oppositehypotenuse sin G 152 hypotenuse 3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. . sin G, and cos G. Write your answers in simplest terms. 12 16.383041 16 Solve for height. Use a calculator. feet tall. 1.5 Examples. 1 Writing Sine and Cosine Ratios Use the triangle to find sin T, cos T,35 cos 35° C ⬔A (90 x)⬚ ? long and hypotenuse 4.0 units long. Find the measures of its acute angles to the nearest degree. © Pearson Education, Inc., publishing as Pearson Prentice Hall. An example of a trigonometric identity is sin x cos 20 The flagpole is about adjacent to An identity is an equationthat is true for all allowed values of the variable. opposite 20 Use the cosine ratio. 20 height 3 Using the Inverse of Cosine and Sine A right triangle has a leg 1.5 units opposite leg adjacent to lA hypotenuse This can be abbreviated cos A All rights reserved. The cosine of ⬔A is the ratio of the length of theleg adjacent to ⬔A to the hypotenuse. All rights reserved. hypotenuse. sine of ⬔A height cos 35 The sine of ⬔A is the ratio of the length of the leg opposite ⬔A to the length of the B 4.0 Use the inverse of the cosine function to find m⬔A. 1.5 cos A m⬔A ⫺1 4.0 cos1 ( ) 67.975687 0.375 m⬔A 0.375 0.37568 Use the cosine ratio. Use the inverse of the cosine. Use a calculator. Round to the nearest degree. To find m⬔B, use the fact that the acute angles of a right triangle are complementary . m⬔A m⬔B 90 Definition of complementary angles m⬔B 90 Substitute. m⬔B 22 Simplify. 68 The acute angles,rounded to the nearest degree, measure 5 Daily Notetaking Guide Daily Notetaking Guide 68 and 22 . Geometry Lesson 8-4 153 All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Quick Check. Lesson 8-5 1. a. Write the sine and cosine ratios for ⬔X and ⬔Y. X 48 48 64 sin X ⫽ 64 80, cos X ⫽ 80,

sin Y ⫽ 80, cos Y ⫽ 80 Angles of Elevation and Depression NAEP 2005 Strand: Measurement Lesson Objective 1 48 80 Z 64 Topic: Measuring Physical Attributes Use angles of elevation and depression to solve problems Local Standards: ____________________________________ Y Vocabulary. Anangle of elevation is the angle formed by a horizontal line and the line of sight to an sin X ⫽ cos Y when lX and lY are complementary. All rights reserved. All rights reserved. object above the horizontal line. b. In general, how are sin X and cos Y related? Explain. An angle of depression is the angle formedby a horizontal line and the line of sight to an object below the horizontal line. Horizontal line Angle of B depression Horizontal line 2. In Example 2, suppose that the angle the wire makes with the ground is 50°. What is the height of the flagpole to the nearest foot? 20 ft Examples. 3. Find the value of x.Round your answer to the nearest degree. a. b. 10 6.5 © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. 50 10 x x 27 68 41 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 15 feet Angle of elevation A 1 Identifying Angles of Elevation and DepressionDescribe ⬔1 and ⬔2 as they relate to the situation shown. 1 2 3 4 One side of the angle of depression is a horizontal line. ⬔1 is the angle of depression from the airplane to the building . One side of the angle of elevation is a horizontal line. ⬔2 is the angle of elevation 154 Geometry Lesson 8-4Name_____________________________________ Class____________________________ Date ________________ Use the ? 200 tan 35⬚ ratio. Theodolite 5 ft 35 200 ft Use a calculator. b. ⬔4 140 feet tall. ft 140 The building is about 5 ft, or 145 ft tall. All rights reserved. The angle of elevation fromthe person on the ground to the building To find the height of the building, add the height of the theodolite, which is 5 3 Aviation An airplane flying 3500 ft above the ground begins a 2 descent to land at the airport. How many miles from the airport is the airplane when it starts its descent? (Note: The angleis not drawn to scale.) 2 x 3,500 ft You x 2 3500 5280 The airplane is about 156 19 3,500 x Use the sine ratio. 3500 Solve for x. sin 2⬚ 100287 .9792 18.993935 Use a calculator. Divide by 5280 to convert feet to miles. miles from the airport when it starts its descent. Geometry Lesson 8-5 All-In-OneAnswers Version A Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. sin 2⬚ 2. You sight a rock climber on a cliff at a 32 angle of elevation. The horizontal ground distance to the cliff is 1000 ft. Find the line-of-sight distance to the rock climber. Rock climber Ground100287.9792 155 The angle of depression from the building to the person on the ground All rights reserved. x Geometry Lesson 8-5 Solve for x. 140 .041508 35 200 tangent 200 x . 1. Describe each angle as it relates to the situation in Example 1. a. ⬔3 © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. tan 35 airplane Quick Check. x x to the Name_____________________________________ Class____________________________ Date ________________ Top of building with a 5-ft tall theodolite. The angle of elevation tothe top of the building is 35. How tall is the building? Use a diagram to represent the situation. building Daily Notetaking Guide Daily Notetaking Guide 2 Surveying A surveyor stands 200 ft from a building to measure its height from the 32 1000 ft about 1179 ft 3. An airplane pilot sees a life raft at a 26angle of depression. The airplane’s altitude is 3 km. What is the airplane’s surface distance d from the raft? about 6.8 km Daily Notetaking Guide Geometry Lesson 8-5 Geometry 157 39 Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date________________ Lesson 8-6 y 1 Describing a Vector Describe OM as an ordered pair. NAEP 2005 Strand: Geometry Describe vectors Solve problems that involve vector addition 2 Examples. Vectors Lesson Objectives 1Name_____________________________________ Class____________________________ Date ________________ Give coordinates to the nearest tenth. Use the sine and cosine ratios to find the values of x and y. Topic: Position and Direction Local Standards:____________________________________ Vocabulary and Key Concepts. Use sine and cosine. 80 x 80 ( cos 40 Adding Vectors x All rights reserved. . A vector is any quantity with magnitude (size) and direction. arrow A vector can be represented with an All rights reserved. kx1 1 x2, y1 1 y2 l For ax1, y1 and c x2, y2, a c x x 40 O x cos 40 . ) Solve for the variable. 61.28355545 Use a calculator. Because point M is in the third y 80 y 80( sin 40 y 80 M ) 51.42300878 quadrant, both coordinates are . To the nearest tenth, OM negative y sin 40 k261.3, 251.4l . 2 Describing a Vector Direction A boatsailed 12 mi east and 9 mi south. The trip can be described by the vector k12, 29l. Use distance and direction to describe this vector a second way. The magnitude of a vector is its size, or length. N Draw a diagram for the situation. The initial point of a vector is the point at which it starts. the DistanceFormula To find the distance sailed, use magnitude The initial point of vector ON is the distance from its terminal point O to its k5, 2l notation for the vector is N. The ordered pair . y 4 2 4 2 O N 2 4x d ( 12 0 )2 ( d 144 )2 9 0 81 W 12 mi (0, 0) E 9 mi (12, 9) Simplify. d "225 d . Distance Formula S Simplify.15 All rights reserved. A resultant vector is the sum of other vectors. © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. The terminal point of a vector is the point at which it ends. Take the square root. To find the direction theboat sails, find the angle that the vector forms with the x-axis. 9 tan x 0.75 Use the tangent ratio. 12 x ⫺1 tan1 (0.75) The boat sailed Find the angle whose tangent is 0.75. 36.869898 0.75 15 miles at about Use a calculator. 37 south of east . 2 4 Geometry Lesson 8-6 Daily Notetaking Guide DailyNotetaking Guide Name_____________________________________ Class____________________________ Date ________________ 3 Adding Vectors Vectors v 4, 3 and w 4, 3 are shown at the right. Write the sum of the two vectors as an ordered pair. Then draw s, the sum of v and w. To find thefirst coordinate of s, add the first 2 To find the second coordinate of s, add the second coordinates of 2 v and w. 4 8x 4 6 (4, 3) Vocabulary. , 3 (3) l Add the coordinates. 4 Simplify. 2 Draw vector v using the origin as the initial point. Draw vector w using the terminal point of v, (4, 3), as the initial point. Drawthe resultant vector s from the the terminal point initial point of v to of w. A transformation of a geometric figure is a change in its position, shape, or size. v 2 w 4 s8 x 6 All rights reserved. 44 y All rights reserved. k NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation ofProperties; Position and Direction Local Standards: ____________________________________ s k4, 3l k4, 3l k 8 , 0 l Translations Lesson Objectives 1 Identify isometries 2 Find translation images of figures (4, 3) coordinates of v and w. Lesson 9-1 2 159Name_____________________________________ Class____________________________ Date ________________ y 4 Geometry Lesson 8-6 In a transformation, the preimage is the original image before changes are made. In a transformation, the image is the resulting figure after changes are made.2 An isometry is a transformation in which the preimage and the image are congruent. 4 A translation (slide) is a transformation that maps all points the same distance and in the Quick Check. same direction. y 51 k221.6, 46.2l 65 O x 2. A small airplane lands 246 mi east and 76 mi north of the point fromwhich it took off. Describe the magnitude and the direction of its flight vector. about 257 mi at 17 N of E 3. Write the sum of the two vectors k2, 3l and k24, 22l as an ordered pair. © Pearson Education, Inc., publishing as Pearson Prentice Hall. A composition of transformations is a combination of two ormore transformations. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. Describe the vector at the right as an ordered pair. Give the coordinates to the nearest tenth. Examples. 1 Identifying Isometries Does the transformation appear to be an isometry? Image Preimage The imageappears to be the same as the preimage, but Because the figures appear to be congruent turned . , the transformation appears to be an isometry. k22, 1l 160 40 Geometry Lesson 8-6 Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 9-1 161 All-In-One Answers Version A ©Pearson Education, Inc., publishing as Pearson Prentice Hall. 158 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date ________________ b. List all pairs of corresponding sides. 2 Naming Images and Corresponding Parts In the diagram, 䉭XYZ is an image of 䉭ABC. a. Name the images of ⬔B and ⬔C. C Z NI and SU; ID and UP ; ND and SP Y B Because correspondingvertices of the preimage and the image are listed in the same order, the image of ⬔B is ⬔ Y , A X 3. Find the image of 䉭ABC for the translation (x 2, y 1). and the image of ⬔C is ⬔ Z . A(3, 3), B(1, 1), C(2, 0) b. List all pairs of corresponding sides. Because corresponding sides of the preimage and theimage are XZ , BC and YZ . 3 Finding a Translation Image Find the image of 䉭ABC y under the translation (x, y) → (x 2, y 3). A(1, 2) → A( 1 2 , 2 3) A Use the rule (x, y ) → (x 2, y 3) B B(1, 0) → B( 1 2, 0 3 ) All rights reserved. XY , AC and All rights reserved. listed in the same order, the following pairsare corresponding sides: AB and x A⬘(2, 0) © Pearson Education, Inc., publishing as Pearson Prentice Hall. Image Yes; the figures appear to be congruent by a flip and a slide. 2. In the diagram, NID S SUP. a. Name the images of ⬔I and point D. U lU, P N S © Pearson Education, Inc., publishing asPearson Prentice Hall. The rule is (x, y) S Quick Check. All rights reserved. , the horizontal change is 2 ( ⫺4 P (x ⫹ 6, y ⫺ 1) y B ), or 6 , A C 4 2 A 2 x C 2 and the vertical change is 0 1 , or ⫺1 . and C( 2 , 4 ). Yes; the figures appear to be congruent by a flip. 4 2 to describe the translation. Using A(4, 1)and its image The image of 䉭ABC is 䉭ABC with A( 1 , 1 ), B( 3 , 3 ), 1. Does the transformation appear to be an isometry? Explain. a. b. Preimage Preimage Image B describe the translation 䉭ABC S 䉭ABC. You can use any point on 䉭ABC and its image on 䉭ABC C C(0, 1) → C( 0 2 , 1 3 ) Examples.4 Writing a Rule to Describe a Translation Write a rule to 4 . 5 Adding Translations Tritt rides his bicycle 3 blocks north and 5 blocks east of a pharmacy to deliver a prescription. Then he rides 4 blocks south and 8 blocks west to make a second delivery. How many blocks is he now from the pharmacy?The rule (x, y) → (x ⫹ 5, y ⫹ 3) represents a ride of 3 blocks north and 5 blocks east. The rule (x, y) → (x ⫺ 8, y ⫺ 4) represents a ride of 4 blocks south and 8 blocks west. Tritt’s position after the second delivery is the composition of the two translations. ( 0 , 0 ) translates to ( 0 5 , 0 3 ) or ( 5 , 3 ). Then, (5 , 3 ) translates to ( 5 ⫺8 , 3 ⫺4 ) or ( ⫺3 , ⫺1 ). Tritt is 1 block south and 3 blocks of the west pharmacy. I D Geometry Lesson 9-1 Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 9-1 163 Name_____________________________________ Class____________________________Date________________ Name_____________________________________ Class____________________________ Date ________________ Quick Check. Lesson 9-2 4. Use the rule (x, y) S (x 7, y 1) to find the translation image of 䉭LMN. Graph the image 䉭LMN. y 6 Reflections Lesson Objectives1 Find reflection images of figures 2 2 M 2 4 L 4 6 A reflection in line r is a transformation such that if a point A is on line r, then x N M 6 All rights reserved. 2 N Vocabulary. All rights reserved. 6 4 L NAEP 2005 Strand: Geometry Topics: Transformation of Shapes and Preservation of Properties LocalStandards: ____________________________________ 4 the image of A is itself B is the point such that r is the , and if a point B is not on line r, then its image perpendicular bisector of BBr. r A A B Line of 5. Refer to Example 5. Tritt now makes a third delivery. Starting from where the second deliverywas, he goes 5 blocks north and 1 block west. How many blocks is Tritt from the pharmacy? reflection B He is 4 blocks north and 4 blocks west of the pharmacy. Example. 164 Geometry Lesson 9-1 All-In-One Answers Version A Daily Notetaking Guide © Pearson Education, Inc., publishing as PearsonPrentice Hall. 1 Finding Reflection Images If point Q(1, 2) is reflected across © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 162 line x 1, what are the coordinates of its reflection image? y x=1 Q Q is 2 units to the left of thereflection line, so its image Q is 2 units to the right of the reflection line. The reflection line is the x 2 2 perpendicular bisector of QQr if Q is at (3, 2) . Quick Check. 1. What are the coordinates of the image of Q if the reflection line is y 1? (1, 4) Daily Notetaking Guide Geometry Lesson 9-2 Geometry 165 41Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________Examples. Lesson 9-3 2 Drawing Reflection Images 䉭XYZ has vertices X(0, 3), Y(2, 0), y and Z(4, 2). Draw 䉭XYZ and its reflection image in the line x 4. First locate vertices X, Y, and Z and draw 䉭XYZ in a coordinate plane. x4 ‚ 4 X 2 O 2 of XXr, YYr, and ZZr. Y 6 Topic: Transformation of Shapes andPreservation of Properties Local Standards: ____________________________________ 8x W , PD and P PDr and PW also form congruent angles with line ᐉ All rights reserved. A rotation of x about a point R is a transformation for which the following are true: PDr form congruent angles with line ᐉ. 艎(that is, R itself • The image of R is • For any point V, RV RV and m⬔VRV x The center of a regular polygon is a point that is vertices from the polygon’s V R R ). R x . V equidistant . angles are congruent. vertical Therefore, PD and PW form congruent angles with line ᐉ by the Draw and identify rotationimages of figures ‚ D Show that PD and PW form congruent angles with line ᐉ. because NAEP 2005 Strand: Geometry Vocabulary. 3 Congruent Angles D is the reflection of D across ᐉ. isometry 1 2 Draw the reflection image XYZ. Because a reflection is an Y ‚ All rights reserved. perpendicular bisector XZ Z Locate points X, Y, and Z such that the line of reflection x 4 is the Rotations Lesson Objective Examples. Property. Transitive 1 Drawing a Rotation Image Draw the image of 䉭LOB under a 60 rotation D about C. Step 1 Use a protractor to draw a 60 angle at vertex C with one side CO. A 4 C C B B 84 4 8 12 x 4 8 3. In Example 3, what kind of triangle is 䉭DPD? Imagine the image of point W reflected across line ᐉ. What can you say about 䉭WPW and 䉭DPD? isosceles; they are similar by SAS Step 3 Locate L⬘ and B⬘ in a similar manner. Then draw 䉭L⬘O⬘B⬘. C O O L L L B B C O O LName_____________________________________ Class____________________________ Date________________ A into six equilateral triangles. a. Name the image of B for a 240 rotation about M. B Because 360 6 60 , each central angle of ABCDEF measures 60° . A 240 counterclockwise rotationabout center M moves point B across triangles. The image of point B is point D . four L 167 Geometry Lesson 9-3 3 Finding an Angle of Rotation A regular 12-sided polygon can be formed by stacking congruent square sheets of paper rotated about the same center on top of each other. Find the angle ofrotation about M that maps W to B. Consecutive vertices of the three squares form the outline of a regular 12-sided polygon. M D 360 12 b. Name the image of M for a 60° rotation about F. 䉭AMF is equilateral, so ⬔AFM has measure 180 3 B L .Examples. C E B O LName_____________________________________ Class____________________________ Date ________________ B F B O L C O Daily Notetaking Guide Daily Notetaking Guide 2 Identifying a Rotation Image Regular hexagon ABCDEF is divided B C O L Geometry Lesson 9-2 C O 60° B 166 C O30 , so each vertex of the polygon is a 30⬚ B W M rotation K about point M. 60 . A 60 60 You must rotate counterclockwise through 7 vertices to map point W to image of M under a 60 rotation about point F is point A . Quick Check. All rights reserved. All rights reserved. rotation of 䉭AMF about point Fwould superimpose FM on FA, so the 1. Draw the image of LOB for a 50 rotation about point B. Label the vertices of the image. point B, so the angle of rotation is 7 ? 30 210 . 4 Compositions of Rotations Describe the image of quadrilateral XYZW for a composition of a 145 rotation and then a 215 rotation,both about point X. Two rotations of 145 and 215 about the same point make a total rotation of 145 215, or 360 . Because this forms a about point X, the image is the preimage XYZW complete rotation . O Quick Check. O L 2. Regular pentagon PENTA is divided into 5 congruent triangles. Name theimage of T for a 144 rotation about point X. P A E E X © Pearson Education, Inc., publishing as Pearson Prentice Hall. B B © Pearson Education, Inc., publishing as Pearson Prentice Hall. L 3. In the figure from Example 3, find the angle of rotation about M that maps B to K. 270° 4. Draw the image of thekite for a composition rotation of two 90 rotations about point K. K T 168 42 Geometry Lesson 9-3 Geometry N Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 9-3 169 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. A Step 2 Use a compassto construct COr CO. All rights reserved. 2. 䉭ABC has vertices A(3, 4), B(0, 1), and C(2, 3). Draw 䉭ABC and its reflection image in the line x 3. y 8 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Quick Check. Geometry:All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 9-4 Examples. Symmetry Lesson Objective 1 Name_____________________________________Class____________________________ Date ________________ 1 Identifying Lines of Symmetry Draw all lines of symmetry for the isosceles trapezoid. NAEP 2005 Strand: Geometry Identify the type of symmetry in a figure Topic: Transformation of Shapes and Preservation of Properties Draw any linesthat divide the isosceles trapezoid so that half of the figure is a mirror image of the other half. Local Standards: ____________________________________ Vocabulary. Line symmetry is the same as reflectional symmetry. All rights reserved. A figure has reflectional symmetry if there is a reflection thatmaps the figure onto itself. All rights reserved. A figure has symmetry if there is an isometry that maps the figure onto itself. There is one line of symmetry. 2 Identifying Rotational Symmetry Judging from appearance, do the letters V and H have rotational symmetry? If so, give an angle of rotation. Theletter V does not have rotational symmetry because it must be rotated before it is its own image. 360 Line of Symmetry The letter H is its own image after one half-turn, so it has rotational symmetry with a 3 Finding Symmetry A nut holds a bolt in place. Some nuts have square reflectional ) A figure hasrotational symmetry if it is its own image for some rotation of 180⬚ or less. A figure has point symmetry if it has 180⬚ rotational symmetry. The figure is its own image after one half-turn, so it has rotational symmetry with a The figure also has point 180 angle of rotation. symmetry. © Pearson Education,Inc., publishing as Pearson Prentice Hall. All rights reserved. ( © Pearson Education, Inc., publishing as Pearson Prentice Hall. The heart-shaped figure has symmetry. line angle of rotation. 180 faces, like the top view shown below. Tell whether the nut has rotational symmetry and/or reflectionalsymmetry. Draw all lines of symmetry. The nut has a square outline with a circular opening. The square and circle are concentric. The nut is its own image after one quarter-turn, so it has symmetry. rotational 90 The nut has 4 lines of symmetry. 170 Geometry Lesson 9-4 Daily Notetaking Guide DailyNotetaking Guide 171 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Quick Check. Lesson 9-5 1. Draw arectangle and all of its lines of symmetry. Dilations Lesson Objective 1 Locate dilation images of figures NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties Local Standards: ____________________________________ All rights reserved. All rights reserved.Vocabulary. A dilation is a transformation with center C and scale factor n for which the following are true: itself • The image of C is ) (that is, C C ). • For any point R, R is on CR and CR n C CR . R R 1 3 Q 2. a. Judging from appearance, tell whether the figure at the right has rotational symmetry. If so,give the angle of rotation. dilation RrQr is the image of RQ under a with center C and scale factor 3 . Q yes; 180° An enlargement is a dilation with a scale factor greater than 1. b. Does the figure have point symmetry? 3. Tell whether the umbrella has rotational symmetry about a line and/or reflectionalsymmetry. The umbrella has both rotational and reflectional symmetry. © Pearson Education, Inc., publishing as Pearson Prentice Hall. yes © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Geometry Lesson 9-4 A reductionis a dilation with a scale factor less than 1. Examples. 1 Finding a Scale Factor Circle A with 3-cm diameter and center C is a dilation of concentric circle B with 8-cm diameter. Describe the dilation. The circles are concentric, so the dilation has center C . Because the diameter of the dilation image issmaller, the dilation is a reduction Scale factor: . diameter of dilation image diameter of preimage 5 3 8 The dilation is a reduction with center C and scale factor 172 Geometry Lesson 9-4 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide 3 8 . Geometry Lesson 9-5 Geometry173 43 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ 2Scale Drawings The scale factor on a museum’s floor plan is 1 : 200. The Quick Check. length and width of one wing on the drawing are 8 in. and 6 in. Find the actual dimensions of the wing in feet and inches. 1. Quadrilateral JKLM is a dilation image of quadrilateral JKLM. Describe the dilation. 1 Thefloor plan is a reduction of the actual dimensions by a scale factor of 200 . Multiply each dimension on the drawing by 200 to find the actual dimensions. Then write the dimensions in feet and inches. in. 6 in. 200 1200 in. The museum wing measures 133 ft, 100 ft 133 ft, 4 in. by . 100 ft 3 Graphing DilationImages 䉭ABC has vertices A(2, 3), B(0, 4), and C(6, 12). What are the coordinates of the image of 䉭ABC for a dilation with center (0, 0) and scale factor 0.75? The scale factor is 0.75 , so use B(0, 4) → B( the rule (x, y ) → (0.75x, 0.75y) . C(6, 12) → C( , 0.75 0.75 6 4 ) K x K 4 2 L L The dilation is areduction with center (0, 0) and scale factor 1 2. 2. The height of a tractor-trailer truck is 4.2 m. The scale factor for a model A(2, 3) → A(0.75 ? (2) , 0.75 (3) ) 0.75 0 O M M 3 in. 4 All rights reserved. 1600 All rights reserved. 8 in. 200 y J J 2 1 . Find the height of the model to the nearest centimeter. truck is54 8 cm , 0.75 (12) ) , B (4.5, ⫺9) , and C (0, 3) . 3. Find the image of 䉭PZG for a dilation with center (0, 0) and scale factor 12. Draw the reduction on the grid and give the coordinates of the image’s vertices. y 2 1 Z Z 2 P 1P 2 x 1 1 G 2 Geometry Lesson 9-5 G (1, 0) Vertices: P 174 a2 1 , 1 b 2 4 , Z ,and G a 1 , 21b 2 Daily Notetaking Guide Daily Notetaking Guide All rights reserved. (⫺1.5, ⫺2.25) © Pearson Education, Inc., publishing as Pearson Prentice Hall. A © Pearson Education, Inc., publishing as Pearson Prentice Hall. The vertices of the reduction image of 䉭ABC are Geometry Lesson 9-5175 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 9-6 Examples. 1 Composition of Reflections inIntersecting Lines The letter D is Compositions of Reflections NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties reflected in line x and then in line y. Describe the resulting rotation. Find the image of D through a reflection across line x. Find the image of thereflection through another reflection across line y. Local Standards: ____________________________________ © Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson Objectives 1 Use a composition of reflections 2 Identify glide reflections D D Vocabulary and Key Concepts. 43 A D yreflections . Theorem 9-2 A composition of reflections across two parallel lines is a . translation All rights reserved. A translation or rotation is a composition of two All rights reserved. Theorem 9-1 x The composition of two reflections across intersecting lines is a The center of rotation is is twice the angleformed by the intersecting lines 86 clockwise, or Theorem 9-3 is rotated A composition of reflections across two intersecting lines is a rotation at point A . rotation . rotation . the point where the lines intersect , and the angle 274 . So, the letter D counterclockwise, with the center of 2 Finding a GlideReflection Image 䉭ABC has vertices A(4, 5), B(6, 2), composition of at most three . reflections Theorem 9-5: Isometry Classification Theorem There are only four isometries. They are the following: Reflection Translation R R R R R Glide reflection Rotation R R R R A glide reflection is the composition of aglide (translation) and a reflection across a line © Pearson Education, Inc., publishing as Pearson Prentice Hall. In a plane, one of two congruent figures can be mapped onto the other by a © Pearson Education, Inc., publishing as Pearson Prentice Hall. Theorem 9-4: Fundamental Theorem of Isometriesand C(0, 0). Find the image of 䉭ABC for a glide reflection where the translation is (x, y) → (x, y 2) and the reflection line is x 1. y A A A B 6 4 2 C 4 2 C x 1 (4, 5) S (4 0 , 5 2 ), or ( 4 , 7 ) C 2 First, translate 䉭ABC by (x, y) → (x, y 2). B B 4 6 (6,2) S (6 0 , 2 2 ), or ( 6 , 4 ) x (0, 0) S (0 0 , 0 2 ), or ( 0 , 2 )Then, reflect the translated image across the line x 1 . 2 The glide reflection image 䉭ABC has vertices A B (⫺4, 4) , and C (2, 2) (6, 7) , . parallel to the direction of translation. 176 44 Geometry Lesson 9-6 Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 9-6 177 All-In-OneAnswers Version A Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date________________ Quick Check. Lesson 9-7 1. a. Reflect the letter R across a and then b. Describe the resulting rotation. a R 1 b R Answers may vary. The result is a clockwise rotation about point c through an angle of 2mlacb. Tessellations Lesson Objectives 2 NAEP 2005 Strand: Geometry Identifytransformation in tessellations, and figures that will tessellate Identify symmetries in tessellations Topic: Geometry Local Standards: ____________________________________ R c R Answers may vary. The result is a translation twice the distance between ᐉ and m. R All rights reserved. R b. Useparallel lines ᐉ and m. Draw R between ᐉ and m. Find the image of R for a reflection across line ᐉ and then across line M. Describe the resulting translation. All rights reserved. Vocabulary and Key Concepts. Theorem 9-6 Every triangle tessellates. Theorem 9-7 Every quadrilateral tessellates. Atessellation, or tiling, is a repeating pattern of figures that completely covers a plane, 艎 m without gaps or overlaps. Translational symmetry is the type of symmetry for which there is a translation that maps a 2. a. Find the image of 䉭TEX under a glide reflection where the translation is (x, y) → (x 1, y) andthe reflection line is y 2. Draw the translation first, then the reflection. Glide reflectional symmetry is the type of symmetry for which there is a glide reflection that E E 4 T x 2 2 yes; Order of steps does not matter in a glide reflection. X T maps a figure onto itself. 2 X X 6 4 2 O 4 6 y 2 4 6 E © PearsonEducation, Inc., publishing as Pearson Prentice Hall. T © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. b. Would the result of part (a) be the same if you reflected 䉭TEX first, and then translated it? Explain. figure onto itself. y Examples. 1 Determining Figures That WillTessellate Determine whether a regular 15-gon tessellates a plane. Explain. Because the figures in a tessellation do not overlap or leave gaps, the sum of the measures of the angles around any vertex must be a a a (n 180 2 180 ( 15 15 for n. Simplify. is not a factor of 360, a regular 15-gon tessellate aplane. Geometry Lesson 9-7 179 and one adjoining will completely cover the plane without gaps or overlaps. Use arrows to show a translation. 3 Identifying Symmetries in Tessellations List the Starting at any vertex, the tessellation can be mapped onto itself using a has point 180⬚ rotation, so thetessellation symmetry centered at any vertex. All rights reserved. All rights reserved. symmetries in the tessellation. translation The tessellation also has translational b. symmetry, as can be seen by sliding any triangle Quick Check. 1. Explain why you can tessellate a plane with an equilateral triangle. Theinterior angles of an equilateral triangle measure 60⬚. Because 60 divides evenly into 360, it will tessellate. 180 Geometry Lesson 9-7 All-In-One Answers Version A Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. onto a copy of itself along any of the lines. ©Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Substitute 15 2. Identify a transformation and outline the smallest repeating figure in each tessellation below. a. repeating figures and a transformation in the tessellation. octagon) Name_____________________________________ Class____________________________ Date ________________ 2 Identifying the Transformation in a Tessellation Identify the square Use the formula for the measure of an angle of a regular polygon. 2 Daily Notetaking Guide Daily NotetakingGuide Name_____________________________________ Class____________________________ Date________________ A repeated combination of an ) 156 will not Geometry Lesson 9-6 . Check n Because 156 178 360 to see whether the measure of an angle of a regular 15-gon is a factor of 360.Answers may vary. Sample: rotation 3. List the symmetries in the tessellation. line symmetry, rotational symmetry, glide reflectional symmetry, translational symmetry Daily Notetaking Guide Geometry Lesson 9-7 Geometry 181 45 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date________________ Lesson 10-1 1 Finding a Missing Dimension A parallelogram has 9-in. and 18-in. sides. NAEP 2005 Strand: Measurement The height corresponding to the 9-in. base is 15 in. Find theheight corresponding to the 18-in. base. First find the area of the parallelogram using the 9-in. base and its corresponding 15-in height. Topic: Measuring Physical Attributes Local Standards: ____________________________________ A bh Vocabulary and Key Concepts. its and base A h . height b bhTheorem 10-2: Area of a Parallelogram height and the corresponding A Area of a parallelogram A 9 ( A 135 A bh 18 135 18 7.5 Theorem 10-3: Area of a Triangle and the corresponding base A h . height b 1bh 2 A base of a parallelogram is any of its sides. Altitude Base The altitude of a parallelogramcorresponding to a given base is the segment perpendicular to the line containing that base drawn from the side opposite the base. The height of a parallelogram is the length of its altitude. © Pearson Education, Inc., publishing as Pearson Prentice Hall. a for h. in.2 135 Area of a parallelogram h 135Substitute h Divide each side by h Simplify. 2 Finding the Area of a Triangle Find the area of 䉭XYZ. A 5 12bh A 5 12 ( A5 triangle Area of a 30 )( 13 ) 195 30 Substitute X for b and 13 䉭XYZ has area 195 cm2. Y Geometry Lesson 10-1 1. A parallelogram has sides 15 cm and 18 cm. The heightcorresponding to a 15-cm base is 9 cm. Find the height corresponding to an 18-cm base. 7.5 cm 18 cm 15 cm Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ Lesson 10-2 3 StructuralDesign The front of a garage is a square 15 ft on each side ( 15 )( 10.6 ) 79.5 15 ft Find the force of the wind against the front of the garage. Use the formula for force. )( 80 )2 7795.2 7800 Substitute 304.5 for A and the bases. h 1h(b 1 b ) 1 2 2 A for v. Round to the nearest hundred. 7800 The height of atrapezoid is the perpendicular lb against the front Quick Check. 2. Find the area of the triangle. 5 cm 30 cm2 13 cm 12 cm 3. Critical Thinking Suppose the bases of the square and triangle in Example 3 are doubled to 30 ft, but the height of each figure stays the same. How is the force of the wind againstthe building affected? The force is doubled. Leg Geometry Leg Base Examples. 1 Applying the Area of a Trapezoid A car window is 20 in. shaped like the trapezoid shown. Find the area of the window. 18 in. 36 in. A 5 12h(b1 1 b2) ( 18 )( 20 36 504 The area of the car window is 46 b1 h b2 A5 DailyNotetaking Guide Base distance h between the bases. A 5 12 Geometry Lesson 10-1 d2 d1 1d d 2 1 2 A Simplify. An 80 mi/h wind exerts a force of about of the garage. 184 b2 Theorem 10-5: Area of a Rhombus or a Kite diagonals. 80 © Pearson Education, Inc., publishing as Pearson Prentice Hall.304.5 © Pearson Education, Inc., publishing as Pearson Prentice Hall. F b1 The area of a trapezoid is half the product of the height and the sum of The area of a rhombus or a kite is half the product of the lengths of its F 0.004Av 2 ( Local Standards: ____________________________________ Theorem10-4: Area of a Trapezoid 15 ft ft 2 The total area of the front of the garage is 79.5 304.5 225 ft 2. F 0.004 NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes Vocabulary and Key Concepts. 10.6 ft ft 2 225 All rights reserved. 12 All rights reserved. Area of the triangle: 1bh 2 2 183Areas of Trapezoids, Rhombuses, and Kites Lesson Objectives 1 Find the area of a trapezoid 2 Find the area of a rhombus or a kite with a triangular roof above the square. The height of the triangular roof is 10.6 ft. To the nearest hundred, how much force is exerted by an 80 mi/h wind blowing directlyagainst the front of the garage? Use the formula F 0.004Av 2, where F is the force in pounds, A is the area of the surface in square feet, and v is the velocity of the wind in miles per hour. Use the area formulas for rectangles and triangles to find the area of the front of the garage. F Geometry Lesson 10-1Name_____________________________________ Class____________________________ Date________________ Example. 15 Z 30 cm Quick Check. b 31 cm 13 cm Simplify. 9 cm containing that base. bh for h. h The height of a triangle is the length of the altitude to the line Area of the square: for b.. A base of a triangle is any of its sides. 182 18 for A and 18 The height corresponding to the 18-in base is 7.5 in. the product of half © Pearson Education, Inc., publishing as Pearson Prentice Hall. The area of a triangle is 15 Simplify. Use the area to find the height corresponding to the 18-in. base. b bhSubstitute 9 for b and The area of the parallelogram is h . ) 15 135 The area of a parallelogram is the product of a base All rights reserved. The area of a rectangle is the product of All rights reserved. Theorem 10-1: Area of a Rectangle All rights reserved. Find the area of a parallelogram Find the area of atriangle Daily Notetaking Guide ) Area of a trapezoid Substitute 18 for h, 20 for b1 , and 36 for b2 . Simplify. 504 in.2. Geometry Lesson 10-2 185 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 Examples. Areas of Parallelograms and Triangles LessonObjectives 2 Name_____________________________________ Class____________________________ Date ________________ Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ 2 Finding Area Using a Right Triangle Find the area of A Name_____________________________________ Class____________________________ Date ________________ B 11 ft Examples. trapezoid ABCD. First draw an altitude from vertex B to DC that divides trapezoid ABCDinto a rectangle and a right triangle. The altitude will meet DC at point X. 3 Finding the Area of a Kite Find the area of kite XYZW. 13 ft DX 11 ft and XC 16 ft D X A 5 12 C By the Pythagorean Theorem, BX 2 XC 2 BC 2, 52 Taking the square root, BX 12 ( ( A5 )( b2 ) 11 16 . ft. You may remember triple that5, 12, 13 is a Pythagorean b1 A 5 12 h A 5 12 12 144 . Use the trapezoid area formula. ) 12 Substitute 162 for h, 11 16 for b1 , and for b2 . Simplify. A 5 12 ( 6 )( 5 ) Substitute 6 for d1 and 5 for d2. A5 15 Simplify. 4 cm W 15 cm 2. R S To find the area, you need to know the lengths of both diagonals.Draw diagonal SU, and label the intersection of the diagonals point X. 䉭SXT is a right 24 triangle because the diagonals of a rhombus The diagonals of a rhombus bisect each other, so TX ft 13 ft X are perpendicular. U 12 T ft. You can use the Pythagorean triple 5, 12, 13 or the Pythagorean TheoremQuick Check. to conclude that SX 1. Find the area of a trapezoid with height 7 cm and bases 12 cm and 15 cm. SU 94.5 cm 2 A 5 12 2. In Example 2, suppose BX and BC change so that m⬔C 60 while bases and angles A and D are unchanged. Find the area of trapezoid ABCD. 67.5 !3 or 116.9 ft 2 A 512 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Z 3 cm 4 Finding the Area of a Rhombus Find the area of rhombus RSTU. ft 2. 162 The area of trapezoid ABCD is Use the formula for the area of a kite.The area of kite XYZW is All rights reserved. 13 2 All rights reserved. so BX 2 d2 d1 3 cm X XZ 5 d1 5 3 1 3 5 6 and YW 5 d2 5 1 1 4 5 5 ft 11 ft 5 ft. 16 1 cm Y Find the lengths of the diagonals of kite XYZW. Because opposite sides of rectangle ABXD are congruent, 10 d1 ( A5 5 ft. ft because thediagonals of a rhombus bisect each other. d2 24 )( Area of a rhombus ) 10 24 Substitute 120 for d1 and 10 for d2 . Simplify. The area of rhombus RSTU is ft 2. 120 Quick Check. 3. Find the area of a kite with diagonals that are 12 in. and 9 in. long. 54 in 2 4. Critical Thinking In Example 4, explain how youcan use a Pythagorean triple to conclude that XU 5 ft. 5 2 ⫹ 12 2 ⫽ 13 2 Geometry Lesson 10-2 Daily Notetaking Guide Daily Notetaking Guide 187 Geometry Lesson 10-2 Name_____________________________________ Class____________________________ Date________________Name_____________________________________ Class____________________________ Date ________________ Lesson 10-3 2 Finding the Area of a Regular Polygon A library is in the shape Areas of Regular Polygons Lesson Objective 1 Find the area of a regular polygon Local Standards:____________________________________ a2 a p 1 ap 2 All rights reserved. perimeter. All rights reserved. The area of a regular polygon is half the product of the apothem and the A ( 9.0 a2 Theorem 10-6: Area of a Regular Polygon a< 18.0 Radius To the nearest 10 ft 2, the area is Divide 360 by thenumber of sides. The apothem bisects the vertex angle of m⬔1 isosceles the 1 2 3 triangle formed by the radii. m⬔2 1 2 ( 60 ) 30 90 30 m⬔3 180 ( m⬔1 , m⬔2 188 60 Substitute ) 60 30 , and m⬔3 60 for m⬔1. 180 . )( ) 144 for s, and simplify. 18.0 Daily Notetaking Guide Area of a regular Substitute21.7 polygon 144 for a and for p. Simplify. 1560 ft 2. with apothem 8 cm. Leave your answer in simplest radical form. This equilateral triangle shows two radii forming an angle that measures s 360 120. Because the radii and a side form an isosceles triangle, 3 60⬚ angles. the apothem bisects the 120angle, forming two 60 8 cm 60 You can use a 30-60-90 triangle to find half the length of a side. 1s 5 2 1s 5 2 "3 s 5 16 p 5 ns p5 ( 3 )( A 5 12 ap A 5 12 8 A5 longer leg "3 ? shorter leg ?a "3 ? 8 192 s Substitute 8 for a. "3 Multiply each side by 2 . Find the perimeter. "3 16 ) 48 "3 Substitute 3 for n and 16!3 for s, and simplify. 48 !3 for p. Area of a regular polygon 48 "3 "3 ) The area of the equilateral triangle is 60 All-In-One Answers Version A Substitute 8 for n and 3 Applying Theorem 10-6 Find the area of an equilateral triangle ( )( The sum of the measures of the angles of a triangle is Geometry Lesson10-3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. radii drawn. Find the measure of each numbered angle. 2 144 Center 1 Finding Angle Measures This regular hexagon has an apothem and m⬔2 ) A 12 ap 21.7 A 12 1562.4 A Examples. 1 471.25 21.7 Step 3 Find the area A. Apothem60 9.0 ft Solve for a. Find the perimeter, where n the number of sides of a regular polygon. ( ( 9.0 ft Pythagorean Theorem p ns The apothem of a regular polygon is the perpendicular distance from the center to a side. 6 )2 23.5 552.25 Step 2 Find the perimeter p. p (8) The radius of a regular polygon isthe distance from the center to a vertex. )2 ( 81 a2 The center of a regular polygon is the center of the circumscribed circle. 360 23.5 ft Step 1 Find the apothem a. Vocabulary and Key Concepts. m⬔1 18.0 ft of a regular octagon. Each side is 18.0 ft. The radius of the octagon is 23.5 ft. Find the area of thelibrary to the nearest 10 ft 2. Consecutive radii form an isosceles triangle, so an apothem bisects the side of the octagon. To apply the area formula A 12ap, you need to find a and p. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 186 Daily Notetaking Guide Substitute 8 for a and Simplify. 192"3 cm 2. Geometry Lesson 10-3 Geometry 189 47 Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Quick Check. Lesson 10-4 1. At the right, a portion of a regular octagon has radii and an apothem drawn. Find themeasure of each numbered angle. ml1 5 45; ml2 5 22.5; ml3 5 67.5 Perimeters and Areas of Similar Figures Lesson Objective 1 1 NAEP 2005 Strand: Measurement and Number Properties and Operations Find the perimeters and areas of similar figures Topics: Systems of Measurement; Ratios andProportional Reasoning 2 3 Local Standards: ____________________________________ All rights reserved. All rights reserved. Key Concepts. Theorem 10-7: Perimeters and Areas of Similar Figures If the similarity ratio of two similar figures is ba , then 232 cm 2 a b (1) the ratio of their perimeters is (2)the ratio of their areas is 2. Find the area of a regular pentagon with 11.6-cm sides and an 8-cm apothem. a2 b2 and . Examples. 1 Finding Ratios in Similar Figures The triangles at the right are 4 similar. Find the ratio (larger to smaller) of their perimeters and of their areas. 384"3 ft 2 the shortest side ofthe right-hand triangle has length 5 4 the perimeters is 7.5 . . , and the ratio of the areas is 52 42 , or 25 16 Because the ratio of the lengths of the corresponding sides of the regular octagons is 83, the ratio of their areas is 64 320 A 82 32 , or 64 9 . Write a proportion. 2880 A 45 Cross-Product Use theDivide each side by 45 Property. 64 . ft 2. Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ . corresponding sides of two regular octagons is 83. The area of the larger octagon is 320 ft 2. Find the area of thesmaller octagon. All regular octagons are similar. The area of the smaller octagon is Daily Notetaking Guide 5 6.25 2 Finding Areas Using Similar Figures The ratio of the length of the 9 Geometry Lesson 10-3 5 6 5 , and By the Perimeters and Areas of Similar Figures Theorem, the ratio of 64 A 190 5 4From larger to smaller, the similarity ratio is 4 All rights reserved. 3. The side of a regular hexagon is 16 ft. Find the area of the hexagon. © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. The shortest side of the left-handtriangle has length Geometry Lesson 10-4 191 Name_____________________________________ Class____________________________ Date ________________ fields. Each dimension of the larger field is 3 12 times the dimension of the Example. smaller field. Seeding the smaller field costs $8.How much money does seeding the larger field cost? 4 Finding Similarity and Perimeter Ratios The areas of two similar pentagons are 32 in.2 and 72 in.2. What is their similarity ratio? What is the ratio of their perimeters? Find the similarity ratio a : b. The similarity ratio of the fields is 3.5 : 1, so the ratio ofthe areas of the 12.25 to 1 . fields is (3.5) 2 : (1) 2, or a2 Because seeding the smaller field costs $8, seeding 12.25 times as much 12.25 land costs $8 . . a2 Quick Check. 1. Two similar polygons have corresponding sides in the ratio 5 : 7. a. Find the ratio of their perimeters. b. Find the ratio of their areas.5:7 All rights reserved. $98 b2 All rights reserved. Seeding the larger field costs b2 a 32 72 16 b 4 6 Take the square root. 3 Perimeters and Areas of Similar Figures Theorem, 3. The similarity ratio of the dimensions of two similar pieces of window glass is 3 : 5. The smaller piece costs $2.50. What shouldbe the cost of the larger piece? $6.94 Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Quick Check. 54 in.2 Geometry 2 the ratio of the perimeters is also 2 : 3 . The area of the larger parallelogram is

96 in.2. Find the area of the smaller parallelogram. 48 Simplify. 36 By the Geometry Lesson 10-4 The ratio of the areas is a 2 : b 2 . The similarity ratio is 2 : 3 . 25 : 49 2. The corresponding sides of two similar parallelograms are in the ratio 34. 192 © Pearson Education, Inc., publishing as PearsonPrentice Hall. 3 Using Similarity Ratios Benita plants the same crop in two rectangular 4. The areas of two similar rectangles are 1875 ft 2 and 135 ft 2. Find the ratio of their perimeters. 5!5 : 3 Daily Notetaking Guide Geometry Lesson 10-4 193 All-In-One Answers Version A Geometry: All-In-One AnswersVersion A (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 10-5 A NAEP 2005 Strand: Measurement 180(sin 36) Substitute for a and p. (cos 36) (sin 36) Simplify. 770.355778 The area is about Local Standards:____________________________________ 18(cos 36) 1620 A Topic: Measuring Physical Attributes Find the area of a regular polygon using trigonometry Find the area of a triangle using trigonometry 2 A 12 Trigonometry and Area Lesson Objectives 1Name_____________________________________ Class____________________________ Date ________________ Use a calculator. 770 ft2. 2 Surveying A triangular park has two sides that measure 200 ft and 300 ft and form a 65 angle. Find the area of the park to the nearest hundred square feet.Key Concepts. Use Theorem 10-8: The area of a triangle is B and the lengths of two sides c . the sine of the included angle A 1 bc(sin A) 2 Area of 䉭ABC a C b All rights reserved. The area of a triangle is one half the product of All rights reserved. Theorem 10-8: Area of a Triangle Given SAS 300 ft of theincluded sine angle. Area 12 side length side length sine of included angle 200 300 sin 65 Area 12 Area sin 65 30,000 65 200 ft Theorem 9-1 Substitute. Substitute. 27189.23361 Use a calculator. ft2. 27,200 The area of the park is approximately Examples. the one half product of the lengths of two sidesand the 1 Finding Area The radius of a garden in the shape of a regular pentagon A⫽1 2ap All rights reserved. formula C . 18 ft 360 Because a pentagon has five sides, m⬔ACB 5 72 . A M B CA and CB are radii, so CA CB . Therefore, 䉭ACM 䉭BCM by , so m⬔ACM 12m⬔ACB the HL Theorem Use theratio to find a. cosine cos 36 a 18 a 18(cos 36) 36 . sine Use the ratio to find AM. 36 AM 18 sin Use the ratio. AM Solve. 18(sin 36) Use AM to find p. Because 䉭ACM 䉭BCM, AB 2 ? AM . Because © Pearson Education, Inc., publishing as Pearson Prentice Hall. Find the perimeter p and apothem a, andthen find the area using the © Pearson Education, Inc., publishing as Pearson Prentice Hall. is 18 feet. Find the area of the garden. Quick Check. 1. Find the area of a regular octagon with a perimeter of 80 in. Give the area to the nearest tenth. 482.8 in.2 2. Two sides of a triangular building plot are 120 ftand 85 ft long. They include an angle of 85. Find the area of the building plot to the nearest square foot. 5081 ft 2 the pentagon is regular, p 5 ? AB . So p 5(2 ? AM ) 10 ? AM 10 ? 18(sin 36) 180(sin 36) . Finally, substitute into the area formula A 12ap. 194 Geometry Lesson 10-5 Daily Notetaking GuideDaily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Circles and Arcs Lesson Objectives 1Find the measures of central angles and arcs 2 Find circumference and arc length R NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes T Local Standards: ____________________________________ TRS is a C B measures of the two arcs. 1 0 0 mABC mAB 1 mBC A Theorem 10-9: Circumference of a Circle C pd or C d and the circumference of the circle r O . RS is a T P minor arc . mRS m⬔RPS RTS is a S P major arc mTRS 180 A semicircle is half of a A minor arc is smaller than mRTS 360 mRS A major arc is greater circle. a semicircle. than a semicircle. . Adjacent arcs arearcs of the same circle that have exactly one point in common. Concentric circles are circles that lie in the same plane and have the same center. Congruent arcs are arcs that have the same measure and are in the same circle or in congruent circles. A Example. r . O 360 0 length of AB semicircle R S T CTheorem 10-10: Arc Length The length of an arc of a circle is the product of the ratio measure of the arc R S P Arc length is a fraction of a circle’s circumference. . 2pr 0 1 1 Finding the Measures of Arcs Find mXY and mDXM in circle C. B 0 mAB ? 2pr 360 A circle is the set of all points equidistant from agiven point called the center. A center of a circle is the point from which all points are equidistant. A radius is a segment that has one endpoint at the center and the other endpoint on the circle. Congruent circles have congruent radii. A diameter is a segment that contains the center of a circle and has bothendpoints on the circle. © Pearson Education, Inc., publishing as Pearson Prentice Hall. p times the diameter The circumference of a circle is All rights reserved. Postulate 10-1: Arc Addition Postulate The measure of the arc formed by two adjacent arcs is the sum of the All rights reserved. Vocabulary andKey Concepts. © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson 10-6 195 Geometry Lesson 10-5 0 0 0 mXY mXD mDY 0 0 mXY m/ XCD mDY 0 mXY 56 40 0 mXY 96 1 0 1 mDXM mDX mXWM 1 mDXM 56 180 1236 mDXM Arc Addition Postulate M The measure of a minor arc is the measure of its corresponding central angle. Substitute. Y 40 Simplify. D 56 C W Arc Addition Postulate X Substitute. Simplify. Quick Check. 1 0 1 1. Use the diagram in Example 1. Find m⬔YCD, mYMW , mMW, and mXMY. 40; 180;96; 264 A central angle is an angle whose vertex is the center of the circle. Circumference of a circle is the distance around the circle. Pi (π) is the ratio of the circumference of a circle to its diameter. 196 Geometry Lesson 10-6 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking GuideGeometry Lesson 10-6 Geometry 197 49 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________Class____________________________ Date________________ Lesson 10-7 Examples. 2 Applying Circumference A circular swimming pool 16 feet in 1 POOL The pool and the fence are concentric circles. The diameter of the pool 4 ft is 16 ft, so the diameter of the fence is 16 4 4 24 ft. Use the formulafor the circumference of a circle to find the length of fencing material needed. 24 ) 75.36 About 76 Use 3.14 to approximate π. Simplify. 1 3 Finding Arc Length Find the length of ADB in circle M in terms of π. 0 Because mAB 150, 1 mADB . Arc Addition Postulate 360 150 210 1 1 pr length of ADB mADBArc Length Postulate 360 ? 2 210 2π 360 ( ) 18 A 150 Quick Check. 2. The diameter of a bicycle wheel is 22 in. To the nearest whole number, how many revolutions does the wheel make when the bicycle travels 100 ft? 17 revolutions 1.3p m joining its endpoints. Sector 1 Applying the Area of a Circle Acircular archery target has a 2-ft diameter. First find the areas of the archery target and the red bull’s-eye. The radius of the archery target is 12 2 ft 1 ft 12 in. ( ) The area of the archery target is πr 2 π area of red region π 100 πr 2 135 424.11501 6 m 100 2 B 10 The area of sector ACB is π π 10π m 2.Quick Check. A 120 24 ft O 0 mAB 360 sector . Substitute. π 192 24 2 h hypotenuse 2 ? shorter leg 12 h Divide each side by 2 . AB !3 ? 2 24 䉭AOB has base © Pearson Education, Inc., publishing as Pearson Prentice Hall. )r 24 ? 576π Simplify. You can use a 30-60-90 triangle to find the height h of 䉭AOB and the length of AB. AB © Pearson Education, Inc., publishing as Pearson Prentice Hall. ? π( 360 1 Use the formula for area of a 2 Step 2 Find the area of 䉭AOB. about 41 in.2 181.5 cm 2 2 ? πr 120 3 1. How much more pizza is in a 14-in.-diameter pizza than in a 12-in. pizza? 2. Critical ThinkingA circle has a diameter of 20 cm. What is the area of a sector bounded by a 208 major arc? Round your answer to the nearest tenth. 199 B area of sector AOB All rights reserved. 5 36 Simplify. Step 1 Find the area of sector AOB. A ) All rights reserved. 424 π in.2 the shaded segment. Round your answerto the nearest tenth. 5 18 in.2 9p Geometry Lesson 10-7 360 in. 3 Finding the Area of a Segment of a Circle Find the area of C π( 6 3 area of yellow region π 9 in.2 144p in. Example. Leave your answer in terms of π. 360 )2 ) 12 (6 The area of the red region is πr 2 π ( 3 )2 The radius of the red bull’s-eyeregion is 12 Name_____________________________________ Class____________________________ Date ________________ 2 Finding the Area of a Sector of a Circle Find the area of sector ACB. area of sector ACB ( Daily Notetaking Guide Name_____________________________________Class____________________________ Date ________________ of a circle It is yellow except for a red bull’s-eye at the center with a 6-in. diameter. Find the area of the yellow region to the nearest whole number. The area of the yellow region is about mAB Segment of a circle Examples. 144 DailyNotetaking Guide B A segment of a circle is the part bounded by an arc and the segment area of archery target 3. Find the length of a semicircle with radius 1.3 m in terms of π. Geometry Lesson 10-6 r O 0 mAB ? pr2 360 intercepted arc. Substitute. A 360 A sector of a circle is a region bounded by tworadii and their D 21 π 1 The length of ADB is 21π cm. 198 measure of the arc . area of the circle. Area of sector AOB 18 cm M A r O The area of a sector of a circle is the product of the ratio B . pr 2 Theorem 10-12: Area of a Sector of a Circle and the ft of fencing material is needed. the product of p andthe square of the radius The area of a circle is All rights reserved. ( Substitute. © Pearson Education, Inc., publishing as Pearson Prentice Hall. C< Theorem 10-11: Area of a Circle All rights reserved. C < 3.14 Vocabulary and Key Concepts. Formula for the circumference of a circle 24 Topic: MeasuringPhysical Attributes 4 ft © Pearson Education, Inc., publishing as Pearson Prentice Hall. Cπ Find the areas of circles, sectors, and segments of circles Local Standards: ____________________________________ 16 ft All rights reserved. pd NAEP 2005 Strand: Measurement 12 "3 12"3 A ( 24"3 144 12"3)( 24 12 ft. 60 60 12 ft ft 30 B ft O "3 ? shorter leg longer leg Multiply each side by 2 . ft A 5 12 bh A 5 12 123 ft 123 A 12 ) "3 12"3 ft, or 24"3 Area of a triangle Substitute 24"3 ft, and height . 12 for b and for h. Simplify. Step 3 Subtract the area of 䉭AOB from the area of sector AOB to find the area of thesegment of the circle. area of segment area of segment< 192 π 353.77047 144 "3 Use a calculator. To the nearest tenth, the area of the shaded segment is 353.8 ft 2. Quick Check. 3. A circle has a radius of 12 cm. Find the area of the smaller segment of the circle determined by a 60 arc. Round youranswer to the nearest tenth. 13.0 cm 2 200 50 Geometry Lesson 10-7 Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 10-7 201 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. diameter will be enclosed in a circular fence 4 ft fromthe pool. What length of fencing material is needed? Round your answer to the next whole number. C Areas of Circles and Sectors Lesson Objective Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________Date________________ Lesson 10-8 Example. Geometric Probability Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ 2 Finding Probability Using Area A circle is inscribed in a square target NAEP 2005 Strand:Data Analysis and Probability Use segment and area models to find the probabilities of events Local Standards: ____________________________________ Find the area of the square. A s2 Vocabulary. 20 2 cm 2 400 Find the area of the circle. Because the square has sides of length 20 cm, the circle’sdiameter is 20 cm, so its radius is 10 cm. Example. 1 Finding Probability Using Segments A gnat lands at random on the All rights reserved. All rights reserved. Geometric probability is a model in which you let points represent outcomes. edge of the ruler below. Find the probability that the gnat lands on apoint between 2 and 10. A πr 2 π ( 10 )2 4 5 6 7 8 9 10 The length of the segment between 2 and 10 is 12 All rights reserved. The length of the ruler is 10 ( A 400 100π ) cm 2 Use areas to calculate the probability that a dart landing randomly in the square does not land within the circle. Use a calculator.Round to the nearest thousandth. area between square and circle P(between square and circle) area of square 2 8 . . length of favorable segment length of entire segment 8 23 12 Quick Check. 1. A point on AB is selected at random. What is the probability that it is a point on CD? A 0 C 1 2 3 4 5 6 7 D B8 9 10 1 π 4 0.215 The probability that a dart landing randomly in the square does not land 21.5 within the circle is about %. Quick Check. 2. Use the diagram in Example 2. If you change the radius of the circle as indicated, what then is the probability of hitting outside the circle? a. Divide the radius by 2.about 80.4% about 96.9% Geometry Lesson 10-8 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ Lesson 11-1 Example. so that it lands entirely within the outer region of the circle atright. Find the probability that this happens with a quarter of radius 15 32 in. Assume that the quarter is equally likely to land anywhere completely inside the large circle. 15 The center of a quarter with a radius of 15 32 in. must land at least 32 in. beyond the boundary of the inner circle in order to lieentirely outside the inner circle. Because the inner circle has a radius of 9 in., the quarter must in., or 9 15 32 9 in. A 5 p° 15 Euler’s Formula 9 32 2 ¢ < 281.66648 in.2 15 in. must land at least Similarly, the center of a quarter with a radius of 32 15 in. within the outer circle. Because the outer circle has aradius of 12 in., 32 15 in., or the quarter must land inside the circle whose radius is 12 in. 32 11 17 32 Local Standards: ____________________________________ Vocabulary and Key Concepts. in. Find the area of the circle with a radius of 915 32 in. pr2 NAEP 2005 Strand: Geometry Topic: Dimensionand Shape 12 in. All rights reserved. 15 32 Space Figures and Cross Sections Lesson Objective 1 Recognize polyhedra and their parts 2 Visualize cross sections of space figures All rights reserved. land outside the circle whose radius is 9 in. in. The numbers of faces (F), vertices (V), and edges (E) of apolyhedron are related by the formula F⫹V⫽E⫹2 . A polyhedron is a three-dimensional figure whose surfaces are polygons. A face is a flat surface of a polyhedron in the shape of a polygon. Find the area of the circle with a radius of 1117 32 in. 32 ¢ < 417.73672 in.2 Use the area of the outer region to findthe probability that the quarter lands entirely within the outer region of the circle. 417.73672 281.66648 P(outer region) area of outer circle area of large circle 417.73672 P(outer region) 136.07024 Rectangular prism Faces 0.32573 417.73672 The probability that the quarter lands entirely within the outerregion of 32.6 0.326 the circle is about or %. Quick Check. 3. Critical Thinking Use Example 3. Suppose you toss 100 quarters. Would you expect to win a prize? Explain. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 11 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall.A pr2 5 p ° 17 203 Geometry Lesson 10-8 Name_____________________________________ Class____________________________ Date ________________ 3 Applying Geometric Probability To win a prize, you must toss a quarter © Pearson Education, Inc., publishing as Pearson Prentice Hall.100π b. Divide the radius by 5. 2 5 202 400 P(landing between 2 and 10) 400 11 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2 cm 2 100π Find the area of the region between the square and the circle. 1 20 cm with 20-cm sides. Find the probability that a dart landing randomly within the square does not land within the circle. Topic: Probability An edge is a segment that is formed by the intersection of two faces. Edge Vertex A vertex is a point where three or more edges intersect. A cross section is the intersection of asolid and plane. Examples Yes; theoretically you should win 32.6 times out of 100. 204 Geometry Lesson 10-8 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-1 Geometry 205 51 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Examples. Example. 1 Identifying Vertices, Edges, and FacesHow many vertices, edges, and faces are there in the polyhedron shown? Give a list of each. There are four There are six There are four A, B, C, and D vertices: the solid in Example 1. A Count the regions: F 4 . AB, BC, CD, DA, AC, and BD edges: 3 Verify Euler’s Formula for the two-dimensional net ofB . Count the vertices: V 6 faces: kABC, kABD, kACD, and kBCD . C Count the segments: E 9 D 4 6 9 1 FVE2 Euler’s Formula 6 8 E 2 Substitute the number of faces and vertices. E 12 All rights reserved. edges on a solid with 6 faces and 8 vertices. Simplify. A solid with 6 faces and 8 vertices has Allrights reserved. 2 Using Euler’s Formula Use Euler’s Formula to find the number of Quick Check. 3. The figure at the right is a trapezoidal prism. a. verify Euler’s Formula F V E 2 for the prism. edges. 12 6 8 12 2 Quick Check. 1. List the vertices, edges, and faces of the polyhedron. T U V 2. Use Euler’sFormula to find the number of edges on a polyhedron with eight triangular faces. 12 edges Geometry Lesson 11-1 Name_____________________________________ Class____________________________ Date________________ 6 14 19 1 207 Geometry Lesson 11-1Name_____________________________________ Class____________________________ Date ________________ Surface Areas of Prisms and Cylinders Lesson Objectives 1 Find the surface area of a prism 2 Find the surface area of a cylinder c. Verify Euler’s Formula F V E 1 for your two-dimensional net. Daily Notetaking Guide Daily Notetaking Guide Lesson 11-2 Possible answer: A prism is a polyhedron with exactly two congruent, parallel prism Pentagonal faces. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes The bases of a polyhedron are are the parallelfaces. Local Standards: ____________________________________ Bases Lateral faces are the faces on a prism that are not bases. Lateral edges An altitude is a perpendicular segment that joins the planes Vocabulary and Key Concepts. of the bases. perimeter of the base height and the All rightsreserved. The lateral area of a right prism is the product of the . L.A. ph lateral area The surface area of a right prism is the sum of the areas of the two bases the and . All rights reserved. Theorem 11-1: Lateral and Surface Area of a Prism a bh c d Base a b h h h Lateral Area Height Lateral Area phSurface Area L.A. 2B Area of a base Theorem 11-2: Lateral and Surface Area of a Cylinder The lateral area of a right cylinder is the product of the circumference of the base height of the cylinder and the . L.A. 2πrh, or L.A. πdh The surface area of a right cylinder is the sum of the areas of the two circularbases lateral area . S.A. L.A. 2B, or S.A. 2πrh 2πr 2 r h Lateral 2πr 52 prism Triangular faces. d d Base c 208 Bases The lateral area of a prism is the sum of the areas of the lateral two bases. c Area Geometry Lesson 11-2 Geometry h Surface Area r A right prism is a prism whose lateral faces arerectangular regions and a lateral edge is an altitude. prisms right © Pearson Education, Inc., publishing as Pearson Prentice Hall. c d © Pearson Education, Inc., publishing as Pearson Prentice Hall. Perimeter of base abcd a b c d h and the Lateral faces The surface area of a prism or a cylinder is the sumof the lateral area and the areas of the S. A. L.A. 2B Perimeter The height is the length of the altitude. h An oblique prism is a prism whose lateral faces are not perpendicular to the bases. h Bases A cylinder is a three-dimensional figure with two congruent circular bases that lie in parallel planes. In acylinder, the bases are parallel circles. h right cylinders The lateral area of a cylinder is the area of the curved surface. A right cylinder is one where the segment joining the centers of the bases is an altitude. h An oblique cylinder is one in which the segment joining the centers r Area of a base B π r2 DailyNotetaking Guide prism oblique of the bases is not perpendicular to the bases. oblique Daily Notetaking Guide cylinder Geometry Lesson 11-2 209 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 206 b. Draw a net for the prism. All rights reserved. S ©Pearson Education, Inc., publishing as Pearson Prentice Hall. R, S, T, U, V; RS, RU, RT, VS, VU, VT, SU, UT, TS; kRSU, kRUT, kRTS, kVSU, kVUT, kVTS © Pearson Education, Inc., publishing as Pearson Prentice Hall. R Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date ________________ Examples. 2 Finding Surface Area of Cylinders A companysells cornmeal and barley in cylindrical containers. The diameter of the base of the 6-in.-high cornmeal container is 4 in. The diameter of the base of the 4-in.-high barley container is 6 in. Which container has the greater surface area? 1 Using Formulas to Find Surface Area Find the surface area of a 10-cmhigh right prism with triangular bases having 18-cm edges. Round to the nearest whole number. L.A. ⫽ ph Use the formula Find the surface area of each container. Remember that r d2. to find the lateral area and Cornmeal Container S.A. ⫽ L.A. ⫹ 2B to find the surface area of the prism. The area B ofthe base is 12 ap, where a is the apothem and p is the perimeter . the formula Substitute the formulas for lateral area of a cylinder and area of a circle. ( )( 6 ) 2π ( 2 2) All rights reserved. All rights reserved. 5 24 p 1 8 5 32 p S.A. 5 L.A. 1 2 Use the formula for surface area. r2 2π 2 cm. 54 B p 5 2prh 1 2cm 18 18 cm The triangle has sides of length 18 cm, so p 3 ? 18 cm, or Barley Container S.A. 5 L.A. 1 2 5 2 p 2π Substitute for r and h. ( 3 )( 4 ) 2π ( 3 2) Simplify. 5 24 p 1 Combine like terms. 5 42 p p Because 42π in.2 ⬎ 32π in.2, the surface area. B h 2pr2 r p 18 container has the greater barley Usethe diagram of the base. 60 a All rights reserved. 1. Use formulas to find the lateral area and surface area of the prism. Divide both sides by "3 9 a "3 ? "3 "3 ? B 12 ? 3 54 9 !3 "3 "3 81 81 !3 )( 54 540 10 ) 2( !3 81 162 !3 cm2. Substitute ph for L.A. ) Substitute 54 for p, 10 81 !3 for h and for B. 400p cm2 3.The company in Example 2 wants to make a label to cover the cornmeal container. The label will cover the container all the way around, but will not cover any part of the top or bottom. What is the area of the label to the nearest tenth of a square inch? 75.4 in.2 Use a calculator. 821 cm 2 Rounded to thenearest whole number, the surface area is 210 for p. 2. Find the surface area of a cylinder with height 10 cm and radius 10 cm in terms of π. Simplify. 820.59223 < 54 for a and Use the formula for surface area. 2B ph Rationalize the denominator. Simplify. S.A. 5 L.A. 1 2B ( "3 3 81 !3 Substitute GeometryLesson 11-2 . Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ Lesson 11-3 Geometry Lesson 11-2 A right cone is a cone in which the altitude is perpendicular to the Surface Areas ofPyramids and Cones Lesson Objectives 1 Find the surface area of a pyramid 2 Find the surface area of a cone 211 Name_____________________________________ Class____________________________ Date ________________ Slant height plane of the base. NAEP 2005 Strand: MeasurementTopic: Measuring Physical Attributes Altitude The slant height of a cone is the distance from the vertex to the edge of the base. Local Standards: ____________________________________ / h r The lateral area of a cone is the area of the curved lateral surface. Base Vocabulary and Key Concepts. Thesurface area of a cone the sum of the lateral area and the area ᐉ . B The surface area of a regular pyramid is the sum of the lateral area S.A. L.A. area of the base and the Examples. pyramid with base edges 7.5 ft and slant height 12 ft. 12 ft B circumference of the base L.A. 12 ? 2πr ? ᐉ, or L.A. and theᐉ slant height . r B prᐉ The surface area of a right cone is the sum of the lateral area and the area of the base. B A regular pyramid is a pyramid whose base is a regular polygon and whose lateral faces are congruent isosceles triangles. The altitude of a pyramid or a cone is the perpendicular segmentfrom the vertex to the plane of the base. The height of a pyramid or a cone is the length of the altitude. Lateral Vertex edge Lateral face Altitude The slant height of a regular pyramid is the length of the altitude of a lateral face. Base Base The lateral area of a pyramid is is the sum of the edge RegularPyramid 7.5 ft The perimeter p of the square base is 4 You are given ᐍ 12 find the lateral area. 1 L.A. 5 L.A. 5 12 ( 30 )( ) Substitute. Simplify. Because the base is a square with side length 7.5 ft, B s2 7.5 2 56.25 S.A. 5 L.A. 1 B The surface area of a pyramid is is the sum of the lateral area and the areaof the base. S.A. 5 180 1 236.25 . Use the formula for surface area of a pyramid. 56.25 Substitute. Simplify. The surface area of the square pyramid is All-In-One Answers Version A Daily Notetaking Guide ft, so you can Find the area of the square base. S.A. 5 Geometry Lesson 11-3 ft. 30 30 Use theformula for lateral area of a pyramid. 12 180 ft, or 7.5 ft and you found that p ᐍ p 2 L.A. 5 areas of the congruent lateral faces. 212 cone 1 Finding Surface Area of a Pyramid Find the surface area of a square . Theorem 11-4: Lateral and Surface Area of a Cone The lateral area of a right cone is half theproduct of the S.A. L.A. All rights reserved. slant height © Pearson Education, Inc., publishing as Pearson Prentice Hall. L.A. and the 1 pᐉ 2 All rights reserved. Theorem 11-3: Lateral and Surface Area of a Regular Pyramid The lateral area of a regular pyramid is half the product of the perimeter of thebase Right of the base. © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 12 m 6m 3 The area of each base of the prism is 432 m 2 ; about 619 m 2 "3 . © Pearson Education, Inc., publishing as Pearson Prentice Hall. 9 B 512ap Quick Check. !3 ? shorter leg longer leg 9 a B . 9 cm © Pearson Education, Inc., publishing as Pearson Prentice Hall. a ? "3 apothem Use 30°-60°-90° triangles to find the 30 9 cm Daily Notetaking Guide 236.25 ft2. Geometry Lesson 11-3 Geometry 213 53 Geometry: All-In-One Answers Version A(continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ 2 Finding Lateral Area of a Cone Leandre usespaper cones to cover her Lesson 11-4 plants in the early spring. The diameter of each cone is 1 ft, and its height is 1.5 ft. How much paper is in the cone? Round your answer to the nearest tenth. 1 2 to find the lateral area of the cone. 0.5 The cone’s diameter is 1 ft, so its radius r is 2 2 1.5 0.25 Theorem11-5: Cavalieri’s Principle If two space figures have the same height and the same cross sectional area "2.5 2.25 Find the lateral area. L.A. 5 p ( L.A. 5 p r ᐍ 0.5 ) Use the formula for lateral area of a cone. "2.5 2.4836471 L.A. < 0.5 Substitute "2.5 for r and All rights reserved. 0.5 for ᐍ. height of the prismheight of the cylinder © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2. Find the lateral area of a cone with radius 15 in. and height 20 in. to the nearest square inch. 1178 in.2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. V Bh, or V 55 m 2 Geometry Lesson 11-3Name_____________________________________ Class____________________________ Date ________________ 2 Finding Volume of a Triangular Prism Find the volume of the prism. A composite space figure is a three-dimensional figure that is the combination of two or more simpler figures.Examples. 1 Finding Volume of a Cylinder Find the volume of the cylinder. Leave your answer in terms of π. The formula for the volume of a cylinder is The diagram shows h and d, but you must find r. altitude and the other leg is the r 5 12d 5 8 V5 p r2 h 5p? 82 ? 9 576 p Use the formula for the volume ofa cylinder. 2 20 841 ( 400 )( 21 The area B of the base is 21bh 12 20 area of the base to find the volume of the prism. V B h V 210 V 8400 ) Simplify. 40 215 Geometry Lesson 11-4 10 m 21 . Use the Simplify. 8400 ft3. 5m 441 Substitute. The volume of the triangular prism is 576π 150 m 3 Use the formulafor the volume of a prism. ? 16 ft Substitute. 6m 210 9 ft 20 m All rights reserved. 2 29 . 1. Find the volume of the triangular prism. 40 m . Use the Pythagorean Theorem to calculate the length of the other leg. V pr 2h Quick Check. 29 m where one leg All rights reserved. base is the h r BName_____________________________________ Class____________________________ Date ________________ with triangular bases. The right triangle area of the base . Daily Notetaking Guide Daily Notetaking Guide right triangular prism B Volume is the space that a figure occupies. 5 base ofthe triangular prism is a h pr 2h The volume of the cylinder is The prism is a area of a base Bh and the 1. Find the surface area of a square pyramid with base edges 5 m and slant height 3 m. . . Theorem 11-7: Volume of a Cylinder The volume of a cylinder is the product of the Quick Check. 214 volumeTheorem 11-6: Volume of a Prism The volume of a prism is the product of the V ft2. 2.5 at every level, then they have the same and the Use a calculator. The lateral area of the cone is about Local Standards: ____________________________________ Vocabulary and Key Concepts. All rights reserved.Topic: Measuring Physical Attributes ft. The altitude of the cone, radius of the base, and slant height form a right Pythagorean Theorem triangle. Use the to find the slant height ᐍ. ᐉ Find the volume of a prism Find the volume of a cylinder 2. The cylinder shown is oblique. a. Find its volume in terms of π.256p m3 16 m m3. 8m 3 Finding Volume of a Composite Figure Find the volume 25 4 cm Bh . ( 14 Volume of Prism II Bh ( 6 Volume of Prism III Bh ( 6 1400 Volume of Prism I Bh 1400 ? ) ? 25 4 ) ? 25 4 ) ? 25 4 ? ? 600 The volume of the composite space figure is 216 54 10 cm 6 cm 4 cm Each prism’svolume can be found using the formula V Sum of the volumes 4 cm 6 cm 6 cm 6 cm 6 cm 4 4 cm Geometry Lesson 11-4 Geometry 2600 600 600 600 © Pearson Education, Inc., publishing as Pearson Prentice Hall. cm 4 cm cm cm 25 14 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 14cm to find the volume. 6 cm rectangular prisms 10 cm You can use three cm of the composite space figure. All rights reserved. prᐍ NAEP 2005 Strand: Measurement b. Find its volume to the nearest tenth of a cubic meter. 804.2 m 3 3. Find the volume of the composite space figure. 2 in. 12 in.3 4 in. 2 in.4 in. 2600 1 in. cm3. Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-4 217 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. Use the formula L.A. Volumes of Prisms and Cylinders Lesson Objectives Geometry: All-In-One Answers VersionA (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 11-5 2 Finding Volume of a Pyramid Findthe volume of a Volumes of Pyramids and Cones square pyramid with base edges 16 m and slant height 17 m. Lesson Objectives 1 Find the volume of a pyramid Find the volume of the cone 2 17 m NAEP 2005 Strand: Measurement The altitude of a right square pyramid intersects the base at the centerof the square. Because each side of the square base is 16 m, the leg of the right triangle along the base is 8 m, as shown. Topic: Measuring Physical Attributes Local Standards: ____________________________________ h 16 m Key Concepts. 16 m Theorem 11-8: Volume of a Pyramid . h B Theorem11-9: Volume of a Cone All rights reserved. height of the pyramid and the 1Bh 3 All rights reserved. area of the base V 17 m h The volume of a pyramid is one third the product of the 8m Step 1 Find the height of the pyramid. The volume of a cone is one third the product of the and the area of the base V13Bh, or V . height of the cone h 1pr 2h 3 172 5 82 1 h2 Use the 289 5 64 1 h2 Simplify. 5 h2 225 r h5 B Pythagorean Theorem 64 Subtract 15 . from each side. Find the square root of each side. Examples. All rights reserved. 1 Finding Volume of a Pyramid Find the volume of a square pyramid with baseedges 15 cm and height 22 cm. Because the base is a square, B ? 15 225 . 1 V5 13 15 B h Use the formula for volume of a pyramid. 3 ( V5 225 )( 22 ) 1650 Substitute 225 for B and 22 for h. Simplify. The volume of the square pyramid is 1650 cm3. © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Step 2 Find the volume of the pyramid. 1 V5 B ( 5 13 5 16 ? 15 Substitute. 1280 Simplify. 1280 m 3. of the oblique cone in terms of π. r5 V5 1 d 5 3 in. 2 11 in. 1 p 3 r2 33 h 11 ) p Use the formula for the volume of a cone.Substitute 3 for r and 11 6 in. for h. Simplify. The volume of the cone is 33p in.3. Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-5 219 Name_____________________________________ Class____________________________ Date ________________Name_____________________________________ Class____________________________ Date________________ Quick Check. Lesson 11-6 1. Find the volume of a square pyramid with base edges 24 m and slant height 13 m. Surface Areas and Volumes of Spheres Lesson Objective 1 Find thesurface area and volume of a sphere 960 m 3 NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes Local Standards: ____________________________________ Vocabulary and Key Concepts. All rights reserved. 12 m 6m All rights reserved. Theorem 11-10: Surface Area of a Sphere2. Find the volume of each cone in terms of π and also rounded as indicated. a. to the nearest cubic meter The surface area of a sphere is four times the product of p and the square S.A. of the radius r of the sphere. 4pr 2 Theorem 11-11: Volume of a Sphere The volume of a sphere is four thirds theproduct of p and 144p m3; 452 m3 the cube S.A. of the radius r of the sphere. 4pr 3 3 A sphere is the set of all points in space equidistant from a 42 mm 21 mm 6174p mm3; 19,396 mm3 3. A small child’s teepee is 6 ft tall and 7 ft in diameter. Find the volume of the teepee to the nearest cubic foot. 77 ft 3© Pearson Education, Inc., publishing as Pearson Prentice Hall. b. to the nearest cubic millimeter © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. ) 3 Finding Volume of an Oblique Cone Find the volume 5 Geometry Lesson11-5 16 The volume of the square pyramid is 5 13 p( 3 2)( 218 Use the formula for the volume of a pyramid. h 3 center diameter given point. The center of a sphere is the given point from which all points r on the sphere are equidistant. The radius of a sphere is a segment that has one endpoint at d thecenter and the other endpoint on the sphere. The diameter of a sphere is a segment passing through the center with endpoints on the sphere. circumference sphere A great circle is the intersection of a plane and a sphere containing the center of the sphere. A great circle divides a sphere into twocongruent Hemispheres Great hemispheres. circle The circumference of a sphere is the circumference of its great circle. 220 Geometry Lesson 11-5 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-6 Geometry 221 55 Geometry: All-In-One AnswersVersion A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date ________________ Examples. Example. 1 Finding SurfaceArea The circumference of a rubber ball is 13 cm. 3 Using Volume to Find Surface Area The volume of a sphere is 1 in.3. Find its surface area to the nearest tenth. Calculate its surface area to the nearest whole number. Step 1 First, find the radius. 13 13 2p p r 5 2pr Substitute 5r Solve for r. 13 for C. 3 5r2 5 4p ? ( Use the formula for surface area of a sphere. ) 13 13 2 2p Substitute for r. 53.794371 Use a calculator. cm2. Substitute r p 30 2 15 . Simplify. The volume of the sphere is cm3. 4500p Quick Check. 1. Find the surface area of a sphere with d 14 in. Give your answer in two ways, in terms of π androunded to the nearest square inch. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 15 4500 Use the formula for volume of a sphere. © Pearson Education, Inc., publishing as Pearson Prentice Hall. r3 3 m 30 c 5 p r cube root Find the of each side. 4p 4.83597587 Use r, S.A. 4π r 2, and acalculator. in.2. 4.8 Geometry Lesson 11-6 100 in.2 3. Find the volume to the nearest cubic inch of a sphere with diameter 60 in. 113,097 in.3 4. The volume of a sphere is 4200 ft 3. Find the surface area to the nearest tenth. 1258.9 ft 2 196p in.2, 616 in.2 222 Solve for r 3. 2. Find the surface area of aspherical melon with circumference 18 in. Round your answer to the nearest ten square inches. Leave your answer in terms of π. 5 43p ? Substitute. Quick Check. 54 2 Finding Volume Find the volume of the sphere. 4 Use the formula for volume of a sphere. r3 S. A. To the nearest whole number, thesurface area of the rubber ball is 3 r3 p To the nearest tenth, the surface area of the sphere is Simplify. p < 3 3 2p 169 5 V5 All rights reserved. All rights reserved. 4p p 3 3 1 4 3πr Step 2 Use the radius to find the surface area. S.A. 5 4 4 V5 Use the formula for circumference. All rights reserved. C5 2Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 11-6 223 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date________________ Lesson 11-7 2 Finding the Similarity Ratio Find the similarity ratio of two similar Areas and Volumes of Similar Solids cylinders with surface areas of 98π ft 2 and 2π ft 2. Use the ratio of the surface areas to find the similarity ratios. NAEP 2005 Strand: Measurement Topic: Systems ofMeasurement a2 5 b2 Vocabulary and Key Concepts. (2) the ratio of their volumes is a3 : b3 . All rights reserved. All rights reserved. Theorem 11-12: Areas and Volumes of Similar Solids If the similarity ratio of two similar solids is a : b, then (1) the ratio of their corresponding areas is a2 : b2 , and 98p a25 b2 Local Standards: ____________________________________ a b5 Simplify. 1 7 8 in. 9 in. 3 in. Both figures have the same shape. Check that the ratios of the corresponding dimensions are equal. The ratio of the radii is The cones are 224 56 3 9 , and the ratio of the heights is not similar GeometryLesson 11-7 Geometry because 8 3 9 ⬆ 26 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 26 in. . 48 a3 5 3 b a b5 The ratio of the volumes is 162 : b3 . Simplify. 27 2 Take the cube root of each side. 3 Step 2 Use thesimilarity ratio to find the surface area S1 of the smaller pyramid. 22 S1 S2 5 S1 135 The ratio of the surface areas is 32 S1 S2 5 4 4 Substitute 9 the Daily Notetaking Guide a2 : b2 . Simplify. 9 4 9 60 ? 135 larger 135 for S2, the surface area of pyramid. Solve for S 1 . Simplify. The surface area of thesmaller pyramid is Daily Notetaking Guide a3 8 S1 5 . of each side. 3 Using a Similarity Ratio Two similar square pyramids have volumes of S1 5 8 26 square root Take the 1 The similarity ratio is 7 : 1 . a3 5 3 b similarity ratio. . Step 1 Find the similarity ratio. The similarity ratio of two similar solids is theratio of their corresponding linear dimensions. Examples. b2 : 2p 49 48 cm 3 and 162 cm 3. The surface area of the larger pyramid is 135 cm 2. Find the surface area of the smaller pyramid. Similar solids have the same shape and all of their corresponding parts are proportional. 1 Identifying Similar SolidsAre the two solids similar? If so, give the a2 The ratio of the surface areas is © Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson Objective 1 Find relationships between the ratios of the areas and volumes of similar solids 60 cm2. Geometry Lesson 11-7 225 All-In-One Answers VersionA Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date ________________ QuickCheck. Lesson 12-1 1. Are the two cylinders similar? If so, give the similarity ratio. Tangent Lines Lesson Objectives yes; 6 : 5 1 10 12 6 2 5 NAEP 2005 Strand: Geometry Use the relationship between a radius and a tangent Use the relationship between two tangents from one point Topic: RelationshipsAmong Geometric Figures Local Standards: ____________________________________ Vocabulary and Key Concepts. 2. Find the similarity ratio of two similar prisms with surface areas 144 m 2 and 324 m 2. All rights reserved. All rights reserved. Theorem 12-1 A If a line is tangent to a circle, then theline is perpendicular to the radius drawn P to the point of tangency. * ) B O AB ' OP Theorem 12-2 2:3 A P If a line in the plane of a circle is perpendicular to a radius at its endpoint the line is tangent to the circle on the circle, then * ) . AB is tangent to 䉺 O. O B A B 3. The volumes of two similar solids are128 m 3 and 250 m 3. The surface area of the larger solid is 250 m 2. What is the surface area of the smaller solid? 160 m2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Theorem 12-3 The twosegments tangent to a circle from a point outside the circle congruent are . O AB > CB C A tangent to a circle is a line, segment, or ray in the plane of the circle that intersects the circle in exactly one point. The point of tangency is the point where a circle and a tangent intersect. A tangent B point oftangency A triangle is inscribed in a circle if all vertices of the triangle lie on the circle. A triangle is circumscribed about a circle if each side of the triangle is tangent to the circle. 226 Geometry Lesson 11-7 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________Class____________________________ Date________________ Examples. 180 Triangle Angle-Sum Theorem x 180 Substitute. x 180 Simplify. x 68 Solve. Y 15 cm X O ODWZ. Because OZ is a radius of 䉺 O, OZ 3 cm. supplement of a right Because the radius of 䉺 P is 7 cm, PD 7⫺3⫽4 OD2 PD2OP2 8 ft R angle, triangle. cm. Simplify. 15.524175 Use a calculator to find the square root. The distance between the centers of the pulleys is about 15.5 cm. Quick Check. 1. ED is tangent to 䉺 O. Find the value of x. D E x 52 O 38 Geometry Lesson 12-1 All-In-One Answers Version A tangent Y S DailyNotetaking Guide Z 6 ft 11 ft C X T 7 ft U W UX XR YS RY 8 ft SZ ZT 6 ft 11 ft By Theorem 12-3, two segments tangent to a circle from a point outside the circle are . congruent WU TW 7 ft p XY YZ ZW WX XR RY YS 11 64 SZ Definition of perimeter p ZT TW WU UX 8 8 6 6 7 7 The perimeter is 228 isnot , PA Find the perimeter of XYZW. W Substitute. OP2 Simplify. PO 2 ⫽ PA 2 ⫹ OA 2 4 Circles Inscribed in Polygons 䉺 C is inscribed in quadrilateral XYZW. Pythagorean Theorem 4 2 OP2 Substitute. 194 Because A . Is 䉭OAP a right triangle? 13 2 5 2 cm. ⬔ODP is also a right angle, and 䉭OPD is aOP 12 2 ⱨ 144 P 13 P 15 cm Because opposite sides of a rectangle have the same measure, 241 PA 2 ⫹ OA 2 12 to 䉺 O at A. D Z right angle, 䉭OAP must be a right triangle, and PO 2 ⱨ PA2 OA2 7 cm 3 cm Draw OP. Then draw OD parallel to ZW to form rectangle Because ⬔ODP is the 5 PO 2 22°two circular pulleys, as shown. Find the distance between the centers of the pulleys. Round your answer to the nearest tenth. 15 O For PA to be tangent to 䉺 O at A, ⬔A must be a B 2 Applying Tangent Lines A belt fits tightly around DW 3 cm and OD 䉺 O such that PO 12, and point A is on 䉺 O suchthat PA 13. Is PA tangent to 䉺 O at A? Explain. right All rights reserved. 22 112 A © Pearson Education, Inc., publishing as Pearson Prentice Hall. angle. All rights reserved. m⬔A m⬔B m⬔C right © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing asPearson Prentice Hall. x° Because BA is tangent to 䉺 C, ⬔A must be a Use the Triangle Angle-Sum Theorem to find x. 2 3 Finding a Tangent 䉺 O has radius 5. Point P is outside C value of x. 15 227 Name_____________________________________ Class____________________________ Date________________ Examples. 1 Finding Angle Measures BA is tangent to 䉺 C at point A. Find the 90 Geometry Lesson 12-1 Daily Notetaking Guide 11 Segment Addition Postulate Substitute. Simplify. 64 ft. Geometry Lesson 12-1 Geometry 229 57 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Quick Check. Lesson 12-2 2. A belt fits tightly around twocircular pulleys, as shown. Find the distance between the centers of the pulleys. 1 35 in. 2 14 in. Chords and Arcs Lesson Objectives 8 in. NAEP 2005 Strand: Geometry Use congruent chords, arcs, and central angles Recognize properties of lines through the center of a circle Topic: Relationships AmongGeometric Figures Local Standards: ____________________________________ about 35.5 in. All rights reserved. All rights reserved. Vocabulary and Key Concepts. Theorem 12-4 Within a circle or in congruent circles (1) Congruent central angles have (2) Congruent chords have (3) Congruent arcs

have congruent congruent congruent chords. arcs. central angles. Theorem 12-5 Within a circle or in congruent circles (1) Chords equidistant from the center are 4. 䉺 O is inscribed in 䉭PQR. 䉭PQR has a perimeter of 88 cm. Find QY. 15 cm X P O 12 cm Z 17 cm © Pearson Education, Inc., publishingas Pearson Prentice Hall. No; 4 2 ⫹ 7 2 ⫽ 8 2 Q Y © Pearson Education, Inc., publishing as Pearson Prentice Hall. (2) Congruent chords are equidistant . congruent from the center. Theorem 12-6 In a circle, a diameter that is perpendicular to a chord bisects the chord arcs and its . All rights reserved. 3. InExample 3, if OA 4, AP 7, and OP 8, is PA tangent to 䉺 O at A? Explain. Theorem 12-7 In a circle, a diameter that bisects a chord (that is not a diameter) is perpendicular to the chord. Theorem 12-8 In a circle, the perpendicular bisector of a chord contains the of the circle. center A chord is a segmentwhose endpoints are on a circle. R P Q O Geometry Lesson 12-1 Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date ________________ Examples. Quick Check. A 1 Using Theorem 12-4 In the diagram, radius OX bisects ⬔AOB. 1. If you are given that BC DF in the circles, what can you conclude? What can you conclude? congruent chords. congruent arcs. congruent Ocentral angles have X O QS 7 7 Substitute. A Q QS 14 Simplify. AB QS Chords that are equidistant from the center of a circle are congruent. AB 14 Substitute 7 14 The distance from the center of 䉺 O to PQ is measured along a perpendicular line. 2 ( 16 P r 15 in. O Substitute. 17 Find the square root ofeach side. 17 18 16 x 36 3. Use the circle at the right. M Substitute. Simplify. OM 2 The radius of 䉺 O is 58 8 Use the Pythagorean Theorem. r 232 ) r 2 289 PM 2 18 S A diameter that is perpendicular to a chord bisects the chord. PQ r 2 8 2 15 2 OP 2 7 16 Q 16 in. The distance from O to PQ is 15 in.,and PQ 16 in. Find the radius of 䉺 O. PM R 2. Find the value of x in the circle at the right. for QS. 3 Using Diameters and Chords P and Q are points on 䉺 O. 1 4 C 4 All rights reserved. B Segment Addition Postulate 2 F B QS QR RS PM C chords have 2 Using Theorem 12-5 Find AB. 1 D P 䉺O 䉺P ©Pearson Education, Inc., publishing as Pearson Prentice Hall. because BX congruent 0 0 AX BX because B 0 0 lO O lP; BC O DF by the definition of an angle bisector. All rights reserved. AX lBOX © Pearson Education, Inc., publishing as Pearson Prentice Hall. ⬔AOX 231 Geometry Lesson 12-2 a. Findthe length of the chord to the nearest unit. 6.8 about 11 4 x b. Find the distance from the midpoint of the chord to the midpoint of its minor arc. 2.8 in. Geometry Lesson 12-2 Geometry Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-2 233 All-In-One Answers Version A © PearsonEducation, Inc., publishing as Pearson Prentice Hall. 230 Geometry: All-In-One Answers Version A (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 12-3 Topic: Relationships Among Geometric Figures x LocalStandards: ____________________________________ x x B 1 mAC 2 C All rights reserved. half the measure of its intercepted arc. m⬔B All rights reserved. Theorem 12-9: Inscribed Angle Theorem A Theorem 12-10 The measure of an angle formed by a tangent and a chord is B half the measure ofthe intercepted arc. 1 1 2mBDC m⬔C x An inscribed angle has a vertex that is on a A Intercepted arc circle and sides that are chords of the circle. . B C Inscribed angle An intercepted arc is an arc with endpoints on the sides of an inscribed angle and its other © Pearson Education, Inc., publishing asPearson Prentice Hall. supplementary © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. . angle. 3. The opposite angles of a quadrilateral inscribed in a circle are DE (m 0 2 0 m EF ) points in the interior of the angle. x° 1 ( 80 70 ) 75 1 mEFG 1 y 1 2 1 2 y ( 0 m EFInscribed Angle Theorem ( 70 0 m FG 120 ) Arc Addition Postulate ) Substitute. 95 Simplify. 2 Using Corollaries to Find Angle Measures Find the values of a and b. By Corollary 2 to the Inscribed Angle Theorem, an angle inscribed in a semicircle is a right angle, so a B 䉺 O is 180 . Therefore, the anglewhose intercepted arc has measure b must have measure 180 Theorem 12-10 0 0 0 mRT m URT m UR m⬔BRT 1 2 ( 180 126 27 ) Substitute and 126 180 126° A Use the properties of tangents to find m⬔TRS. m⬔BRS 90 m⬔BRS m⬔ BRT 63 m⬔TRS 27 perpendicular to the radius of a circle at itspoint of tangency. 90 m⬔ m⬔TRS TRS 58 a° . twice the measure of the inscribed 235 Geometry Lesson 12-3 Angle Measures and Segment Lengths NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures Local Standards: ____________________________________ Vocabularyand Key Concepts. U Simplify. A tangent is or S for mURT for mUR. All rights reserved. m⬔BRT N Arc Addition Postulate 32 the measure of the intercepted 116 . Lesson Objectives 1 Find the measures of angles formed by chords, secants, and tangents 2 Find the lengths of segments associated withcircles R 0 mRT All rights reserved. 2 ) 90 half Lesson 12-4 Angle Addition Postulate Substitute. Theorem 12-11 The measure of an angle formed by two lines that (1) intersect inside a circle is half the sum of the measures of the y° 1 m⬔1 1 (x 1 y) 2 x° difference of the measures of the Simplify. x°intercepted arcs. (2) intersect outside a circle is half the y° 1 y° 1 x° y° 1 x° intercepted arcs. m⬔1 Quick Check. 1 (x 2 y) 2 105 2. For the diagram at the right, find the measure of each numbered angle. ml1 5 90, ml2 5 110, ml3 5 90, ml4 5 70 4 1 60° 2 80° 3 3. In Example 3, describe two ways to findm⬔NRS using Theorem 12-10. mlNRS 5 mlNRU 1 mlURS 5 1 2mRU 1 27 5 63 1 27 5 90 0 1 m⬔NRS 1 2m RUS 2(180) 90 236 Geometry Lesson 12-3 All-In-One Answers Version A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. In Example 1, find m⬔DEF. © Pearson Education,Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 O Name_____________________________________ Class____________________________ Date ________________ T RB is tangent to 䉺 A at point R. Find m⬔BRT and m⬔TRS. 32° b°Daily Notetaking Guide Daily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ m⬔BRT 90 . The sum of the measures of the three angles of the triangle inscribed in angle, so b 2( 58 3 *Using ) Theorem 12-10 RS andTU are diameters of circle A. G Simplify. Because the inscribed angle has Geometry Lesson 12-3 90° Substitute. arc, the intercepted arc has 234 D y° Arc Addition Postulate 2 y congruent 1 0 1 Because EFG is the intercepted arc of ⬔D, you need to find mFG in order 1 to find mEFG . The arc measure ofa circle is 360, so 0 mFG 360 70 80 90 120 . y Corollaries to the Inscribed Angle Theorem right C 2 C C 2. An angle inscribed in a semicircle is a Inscribed Angle Theorem B D D 1. Two inscribed angles that intercept the same arc are F 80° 2 Vocabulary and Key Concepts. The measure of an inscribedangle is 1 mDEF 1 70° E 1 Using the Inscribed Angle Theorem Find the values of x and y. NAEP 2005 Strand: Geometry Find the measure of an inscribed angle Find the measure of an angle formed by a tangent and a chord 2 Examples. Inscribed Angles Lesson Objectives 1Name_____________________________________ Class____________________________ Date ________________ Theorem 12-12 For a given point and circle, the product of the lengths of the two segments from the point to the circle is constant along any line through the point and circle. I. a c b P dabcd III. II. x w z y t P (w x)w (y z)y A secant is a line that intersects a circle at two points. (y z)y Daily Notetaking Guide P t2 secant A O Daily Notetaking Guide y z B Geometry Lesson 12-4 Geometry 237 59 Geometry: All-In-One Answers Version A (continued)Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Examples. 1 Finding Arc Measures An advertising agencywants a frontal photo of a “flying Quick Check. A X T Theorem 12-11(2) 2 [( 12 ( 12 72 72 x A ) ) 360 ⫺ x x ] Substitute. 2x 360 Simplify. x 110° 70° 72° All rights reserved. m⬔T 250 Y All rights reserved. 1 0 (mXAY mXY ) 1 1. Find the value of w. w° saucer” ride at an amusement park. The photographerstands at the vertex of the angle formed by tangents to the “flying saucer.” What is the measure of the arc that will be in the photograph? In the 0 diagram, the photographer stands at 1 ) ) point T. TX and TY intercept minor arc XY and major arc XAY. 0 Let mXY x. 1 360 Then mXAY x. 72 180 72 180 Add xto both sides. x 108 Solve for x. 2. Critical Thinking To photograph a 160 arc of the “flying saucer” in Example 1, should you move toward or away from the object? What angle should the tangents form? away; 20° Distributive Property. 3. Find the value of x to the nearest tenth. 20 x arc will be in theadvertising agency’s photo. 108⬚ 14 13.8 16 200 ft the center of the circle. Because the radius is 125 ft, the 125 250 ft. The length of 2 diameter is B x the other segment along the diameter is 250 ft 50 ft, or 200 ft. xx 50 x2 x Theorem 12-12(1) 200 10,000 Multiply. 100 Solve for x. 200 The shortestdistance from point A to point B is 2x or 238 Center b. Critical Thinking Complete the circle with the given center. Then use Theorem 12-12 to find the value of x. Show that your answers to parts (a) and (b) are equal. 16 in.; (8 2 4"3) 8 1 4"3 Name_____________________________________Class____________________________ Date________________ (x 4 )2 (y 1 )2 25 (x 4 )2 (y 1 )2 ( 5 )2 c c h Key Concepts. y Theorem 12-13 2 8 6 4 2 O Examples. 0 )2 (x 8 ) 2 y2 ( "5 )2 5 Substitute ( "5 for r. 8, 0 ) for (h, k) and Simplify. a circle with center (5, 8) that passes through the point (15, 13).First find the radius. Distance Use the 15 5 8 #( )2 ( 13 )2 2 ) 20 21 #( ( )2 Substitute ( 5, 8 Formula to find r. ) for (h, k) and ( 15, 13 ) for (x, y). Simplify. 441 #400 #841 29 Then find the standard equation of the circle with center (5, 8) and radius (x h )2 (y k )2 r 2 (x 5 )2 (y 8 )2 (x 5 ) 2 (y 8 ) 2 240 60 ( 29841 Geometry Lesson 12-5 Geometry 6 8 x 4 Applying the Equation of a Circle A diagram locates a radio tower at 2 Using the Center and a Point on a Circle Write the standard equation of 2( h ) k x y )2 r #( 4 4 Standard form 29 © Pearson Education, Inc., publishing as Pearson Prentice Hall. )] 2 (y 2 2 ©Pearson Education, Inc., publishing as Pearson Prentice Hall. 8 y 4 1 Writing the Equation of a Circle Write the standard equation of a circle with center (8, 0) and radius "5. ( r and the radius is 5 . (⫺4, 1) 6 All rights reserved. x O All rights reserved. (h, k) Relate the equation to the standard form (x h)2 (yk)2 r 2. c k The center is (x, y) r (h, k) and radius r is (x h) 2 (y k) 2 r 2. [x 239 circle with equation (x 4) 2 (y 1) 2 25. Then graph the circle. Local Standards: ____________________________________ (x h )2 (y k )2 r 2 Geometry Lesson 12-4 3 Graphing a Circle Given Its Equation Find the center andradius of the NAEP 2005 Strand: Geometry Topic: Position and Direction The standard form of an equation of a circle with center 16 1.07 8 1 4"3 Name_____________________________________ Class____________________________ Date ________________ Circles in the Coordinate Plane Lesson12-5 5 Daily Notetaking Guide Daily Notetaking Guide Arc x 8 in. (8 2 4"3) in. ft. Geometry Lesson 12-4 Lesson Objectives 1 Write an equation of a circle 2 Find the center and radius of a circle 4. The basis of a design of a rotor for a Wankel engine is an equilateral triangle. Each side of the triangle is achord to an arc of a circle. The opposite vertex of the triangle is the center of the arc. In the diagram at the right, each side of the equilateral triangle is 8 in. long. a. Use what you know about equilateral triangles and find the value of x. All rights reserved. The perpendicular bisector of the chord AB containsx © Pearson Education, Inc., publishing as Pearson Prentice Hall. A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 50 ft point A to point B along the arc of a circle with a radius of 125 ft. Find the shortest distance from point A to point B. © Pearson Education, Inc., publishing as PearsonPrentice Hall. 2 Finding Segment Lengths A tram travels from (6, 12) on a coordinate grid where each unit represents 1 mi. The radio signal’s range is 80 mi. Find an equation that describes the position and range of the tower. (6, ⫺12) The center of a circular range is at (x h)2 (y k)2 r 2 ( x 6 (x ) 2 [y ( ) ( 62 y 12 12 )] 2 )2 , and the radius is 80 . Use standard form. 80 2 6400 Substitute. This is an equation for position and range of the tower. Quick Check. 1. Write the standard equation of each circle. a. center (3, 5); radius 6 (x ⫺ 3) 2 ⫹ (y ⫺ 5) 2 ⫽ 36 b. center (2, 1); radius "2 (x ⫹ 2) 2 ⫹ (y ⫹ 1) 2 ⫽ 2 .Standard form )2 Substitute ( 5, 8 ) for (h, k) and 29 for r. Simplify. Daily Notetaking Guide Daily Notetaking Guide Geometry Lesson 12-5 241 All-In-One Answers Version A Geometry: All-In-One Answers Version A (continued) Name_____________________________________Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 12-6 2. Write the standard equation of the circle with center (2, 3) that passes through the point (1, 1). Locus:A Set of Points Lesson Objective (x ⫺ 2) 2 ⫹ (y ⫺ 3) 2 ⫽ 13 1 NAEP 2005 Strand: Geometry Draw and describe a locus Topic: Dimension and Shape Local Standards: ____________________________________ 3. Find the center and radius of the circle with equation (x 2) 2 (y 3) 2 100. Then graph thecircle. y Vocabulary. 16 center: (2, 3); radius: 10 x 4 8 All rights reserved. O (2, 3) All rights reserved. A locus is a set of points, all of which meet a stated condition. 8 Examples. 1 Describing a Locus in a Plane Draw and describe the locus of points in a 4. When you make a call on a cellular phone, atower receives the call. In the diagram, the centers of circles O, A, and B are locations of cellular telephone towers. Write an equation that describes the position and range of Tower O. plane that are 1.5 cm from 䉺 C with radius 1.5 cm. Use a compass to draw 䉺 C with radius 1.5 cm. y 24 All rightsreserved. 8 16 20 24 x 8 cm 4 4 5 1. 8 4 O C cm 4 5 1. A 8 © Pearson Education, Inc., publishing as Pearson Prentice Hall. B 16 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 20 x 2 ⫹ y 2 ⫽ 144 The locus of points in the interior of 䉺 C that are 1.5 the center C. cm from 䉺 C is Thelocus of points exterior to 䉺 C that are 1.5 cm from 䉺 C is a circle with center C and radius 3 cm. The locus of points in a plane that are 1.5 cm from a point on 䉺 C with radius 1.5 cm is point C and a circle with radius 3 cm and center C. Geometry Lesson 12-5 Daily Notetaking Guide Daily NotetakingGuide Geometry Lesson 12-6 243 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ 2 Drawing a Locus for TwoConditions Point P is 10 cm from point Q. Quick Check. Describe the locus of points in a plane that are 6 cm from point P and 8 cm from point Q. * ) 1. Draw and describe the locus: In a plane, the points 2 cm from line XY. * ) * ) Two lines parallel to XY , each 2 cm from XY . 6 cm 2 cm 8 cm P Q 10 cm 10X 8 cm center P and radius 6 cm. The set of points in a plane that are 8 cm from point Q is a circle with All rights reserved. The set of points in a plane that are 6 cm from point P is a circle with 2. Draw the locus: In a plane, the points equidistant from two points X and Y and 2 cm from the midpoint of XY.center Q and radius 8 cm. Because the sum (14 cm) of the radii is greater than overlap by 4 10 cm, the circles cm on PQ. The circles intersect at two points, so the locus is the two points where the circles intersect X . 4 cm from a plane M. M 44 cm 艎 © Pearson Education, Inc., publishing as PearsonPrentice Hall. 3 Describing a Locus in Space Describe the locus of points in space that are 44 cm Y 2 cm All rights reserved. 6 cm © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 242 } } 2 cm Y 2 cm 3. Draw and describeeach locus. a. In a plane, the points that are equidistant from two parallel lines. A line midway between the two parallel lines, parallel to and equidistant from each. b. In space, the points that are equidistant from two parallel planes. A plane midway between the two parallel planes, parallel to andequidistant from each. Imagine plane M as a horizontal plane. Because distance is measured along a perpendicular segment from a point to a plane, the locus of points in space that are 4 cm from a plane M are a plane 4 cm above and parallel to plane M and another plane 4 cm below and parallel toplane M. A 244 Geometry Lesson 12-6 All-In-One Answers Version A Daily Notetaking Guide Daily Notetaking Guide B Geometry Lesson 12-6 Geometry 245 61 Geometry: All-In-One Answers Version B (continued) Guided Problem Solving 1-2 Chapter 1 1. They represent three-dimensional objects on atwodimensional surface. 2. nine 3. See the figure in 4, below. Practice 1-1 1. 47, 53 2. 42, 54 3. -64, 128 4. Sample: 2 or 3 5. 6 or 8 6. Y or A 7. any hexagon 8. a 168.75° angle 9. 34 10. Sample: The farther out you go, the closer the 4. ratio gets to a number that is approximately 0.618. 11. 0, 1, 1, 2, 3, 5,8, 13 Guided Problem Solving 1-1 All rights reserved. 1. The pattern is easier to visualize. 2. The graph will go up. 3. Use Years for the horizontal axis. 4. Use Number of Stations for the vertical axis. 5. increasing 6. greater 7. Yes. Since the number of stations increases steadily from 1950 to 2000, we canbe confident that the number of stations in 2010 will be greater than in 2000. 8. Patterns are necessary to reach a conclusion through inductive reasoning. 9. (any list of numbers without a pattern would apply) 2,435; 16,439; 16,454; 3,765; 210,564 5. Yes. It is similar to the foundation drawing, except thereare no numbers. 6. no 7. Practice 1-2 1. Top Fro nt Rig ht nt h Rig 1 1 1 1 1 1 Top Front Right 1 1 1 Front 1. Collinear points lie on the same line. 2. Answers may vary. (Some people might note that the y-coordinate of two of the points is the same so that the third point must have the same y-coordinate tobe collinear. Since it does not, the points are not collinear.) 3. horizontal 4. No. 5. No. 6. All points must have the same y-coordinate, -3. 7. No. 8. a1, 21 b 2 Practice 1-4 1. true 2. false 3. false 4. false 5. JK, HG 6. any * ) 3 Front Top 5. A, C, D 6. C 7. D 8. B 9. A L1 3 Guided Problem Solving 1-3 Front 4. 21. AC 2. any two of the following: ABD, DBC, CBE, ABE, ECD, ADE, ACE, ACD 3. * yes) 4. no 5. yes 6. yes * )7. yes 8. yes 9. G 10. LM * 11.) the empty set 12. KP 13. Sample: plane ABD 14. AB 15. no 16. yes 17. the empty set 18. no 19. yes 20. yes t Right 3. 4 Right Practice * ) 1-3 Right © PearsonEducation, Inc., publishing as Pearson Prentice Hall. 2. Fro Front 8. Yes. 9. All-In-One Answers Version B Right * )* ) * )* ) three of the following pairs: EF and JH ; EF and GK ; HG * )* ) * )* ) * )* ) * )* ) and JE ; HG and FK ; JK and EH ; JK and FG ; EJ and * )* ) * )* ) * )* ) * )* ) * ) FG ; EH and FK ; JE andKG ; EH and KG ; JH and KF ; * ) * ) JH and GF 7. planes A and B 8. planes A and C Geometry 41 Geometry: All-In-One Answers Version B ) ) ) ) ) 9. Sample: EG 10. EF and ED or EG and ED ) ) 11. FE , FD 12. yes 13. Sample: S (continued) Practice 1-6 1. any three of the following: &O, &MOP, &POM,&1 2. &AOB 3. &EOC 4. &DOC 5. 51 6. 90 7. 141 8. 68 9. &ABD, &DBE, &EBF, &DBF, &FBC 10. &ABF, &DBC 11. &ABE, &EBC T Guided Problem Solving 1-6 U 1. Angle Addition Postulate 2. supplementary angles 3. m&RQS + m&TQS = 180 4. (2x + 4) + (6x + 20) = 180 5. x = 19.5 6. m&RQS = 43;m&TQS = 137 7. The sum of the angle measures should be 180; m&RQS + m&TQS = 43 + 137 = 180. 8a. x = 11 8b. m&AOB = 17; m&COB = 73 14. Sample: R Practice 1-7 W All rights reserved. Q 1. P X B Y A 2. Guided Problem Solving 1-4 X 1. Opposite rays are two collinear rays with the sameendpoint. 2. a line 3–4. See graph in Exercise 5 answer. 5. Answers may vary. Sample: (0,0) (Answers will be Y coordinates (x, y), wherey = 32x, x < 2.) y © Pearson Education, Inc., publishing as Pearson Prentice Hall. 3. 6 B(4,6) 4 X A(2,3) 2 Y x 2 2 2 C(0,0) 4 X 6 4. 6. yes 7. L(4, 2) Practice 1-5 1. 4 2.12 3. 20 4. 6 5. 22 6. -3, 4 7. no 8. -2 9. 11 10. 29 11. 29 45° 90° Guided Problem Solving 1-5 1. AD DC 2. AD = DC 3. Segment Addition Postulate. 4. Since AD = DC, AC = 2(AD). 5. AC = 2(12) = 24 6. y = 15 7. DC = AD = 12 8. Answers may vary. 9. ED = 11, DB = 11, EB = 22 42 Geometry All-In-OneAnswers Version B L1 Geometry: All-In-One Answers Version B (continued) Practice 1-8 5. 1.–5. 2 y 6 A 1 C 4 A 2 1 2 6 6. D 4 2 6 x 4 2 E B 4 2 ? 2 2 All rights reserved. 2 6 C 7. X 6. 12 7. 13 8. (5, 5) 9. (-2 12 , 6) 10. (-0.3, 3.4) 11. (5, -2) 12. yes; AB = BC = CD = DA = 6 13. "401 20.025 14. y S 6 Q 8. 42 P © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2 ? X R 2 4 6 8x 15. 24.7 16. (3.5, 3) W Guided Problem Solving 1-8 9. true 10. false 11. false 12. true Guided Problem Solving 1-7 1. &DBC > &ABC 2. complementary angles 3. &CBD 4. m&CBD = m&CBA = 41 5. m&ABD = m&CBA +m&CBD = 41 + 41 = 82 6. m&ABE + m&CBA = 90 m&ABE + 41 = 90 m&ABE = 49 7. m&DBF = m&ABE = 49 8. Answers may vary. Sample: The sum of the measures of the complementary angles should be 90 and the sum of the measures of the supplementary angles should be 180. 9. m&CBD= 21,m&FBD = 69, m&CBA =21, and m&EBA = 69 L1 All-In-One Answers Version B 1. Distance Formula 2. The distance d between two points A(x1,y1) to B(x2,y2) is d 5 Í (x2 2 x1) 2 1 (y 2 2 y1) 2 3. No; the differences are opposites but the squares of the differences are the same. 4. XY 5 Í (5 2 (26)) 2 1 (22 29) 2 5. To the nearest tenth, XY = 15.6 units. 6. To the nearest tenth, XZ = 12.0 units. 7. Z is closer to X. 8. The results are the same; e.g., XY 5 Í (26 2 5) 2 1 (9 2 (22)) 2 5 Í 242 , or about 15.6 units, as before. 9. YZ 5 Í (17 2 (26)) 2 1 (23 2 9) 2 5 Í 673 ; to the nearest tenth, YZ = 25.9 units. To the nearesttenth, XY+YZ+XZ = 53.5 units. Geometry 43 Geometry: All-In-One Answers Version B Practice 1-9 (continued) 1B: Reading Comprehension * ) * )* 1. 792 in.2 2. 2.4 m2 3. 16p 4. 7.8p 5. 26 cm; 42 cm2 6. 29 in.; 42 in.2 7. 40 m; 99 m2 8. 26; 22 9. 30; 44 10. 156.25p 11. 10,000p 12. 36 13. 26; 13 ) * ) 1.Answer may vary. Sample: AB 6 CD , EF 6 GH , Guided Problem Solving 1-9 JK > LM, JL > KM, m/AJF + m/FJK = 180°, ) * ) /HKB > /KMD,DN ' CD 2. Points A, M, and S * ) * ) * ) are collinear. 3. AB , HI , and LN intersect at point M. 4. a 1. six 2. It is a two-dimensional pattern you can fold to form a three-dimensional object. 3. rectangles 1C: Reading/Writing Math Symbols 4. 1. Line BC is parallel to line MN. 2. Line CD 3. Line segment GH 4. Ray AB 5. The length of 6 4 4 4 4 segment XY is greater than the length of segment ST. 6. MN = XY 7. GH = 2(KL) 8. ST ' UV 9. plane ABC || plane XYZ 10. AB 6 DE6 8 4 4 4 1. parallel planes 2. Segment Addition Postulate 3. supplementary angles 4. opposite rays 5. isometric drawing 6. perpendicular lines 7. foundation drawing 8. right angle 9. congruent sides 6 1E: Vocabulary Check 6 2 Net: A two-dimensional pattern that you can fold to form a three-dimensionalfigure. Conjecture: A conclusion reached using inductive reasoning. Collinear points: Points that lie on the same line. Midpoint: A point that divides a line segment into two congruent segments. Postulate: An accepted statement of fact. 2 6. They are equal. 7. 208 in. 8. Answers will vary. Sample: 2(4 6) +2(4 8) + 2(6 8); the results are the same, 208 in.2 9. 6(72) = 294 in.2 5. 208 in. 1A: Graphic Organizer 1. Tools of Geometry 2. Answers may vary. Sample: patterns and inductive reasoning; measuring segments and angles; basic constructions; and the coordinate plane 3. Check students’ work. 44Geometry 1F: Vocabulary Review Puzzle C O L L I N E A R P P D C L Z J P S Y J E Z C C C E O R H L O E G E L K C A F R T S V I L A B B N L I L A E A O R E U N N X W L T Z S L B G N W I N N Y L T I Y Y U R G T A C N E L F Q G G D P C O S C S B R R K A C I E N I L R D M E P I D O U A X TF N R V P V E U A A J D V K C P Y H U E D I C V D N E H X N I I T X S G M S L X S E G M E N T E O M I A V I X K F L P L S D K V T P R C O F J R N E H B L E V D V J S S R R E N L W F Q G N I N O S A E R E V I T C U D N I R O T C E S I B L P J A E N R U K R S T R A I G H T A N G L E W U QA P E L G N A E T U C A W K M F B O Z N T U B A R N K I W T K Q R U I V C F All-In-One Answers Version B © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4 All rights reserved. 1D: Visual Vocabulary Practice 8 L1 Geometry: All-In-One Answers Version B Chapter 2 Practice 2-1 1.Sample: It is 12:00 noon on a rainy day. 2. Sample: 6 3. If you are strong, then you drink milk. 4. If a rectangle is a square, then it has four sides the same length. 5. If x = 26, then x - 4 = 22; true. 6. If m is positive, then m2 is positive; true. 7. If lines are parallel, then their slopes are equal; true. 8.Hypothesis: If you like to shop; conclusion: Visit Pigeon Forge outlets in Tennessee. 9. If you visit Pigeon Forge outlets, then you like to shop. 10. Drinking Sustain makes you train harder and run faster. 11. If you drink Sustain, then you will train harder and run faster. 12. If you All rights reserved. trainharder and run faster, then you drink Sustain. Guided Problem Solving 2-1 1. Hypothesis: x is an integer divisible by 3. 2. Conclusion: x2 is an integer divisible by 3. 3. Yes, it is true. Since 3 is a factor of x, it must be a factor of x ? x = x2. 4. If x2 is an integer divisible by 3 then x is an integer divisible by 3.5. The converse is false. Counterexamples may vary. Let x2 = 3. Then x 5 Í 3 , which is not an integer and is not divisible by 3. 6. No. The conditional is true, so there is no such counterexample. 7. No. By definition, a general statement is false if a counterexample can be provided. 8. If 5x + 3 = 23, then x= 4. The original statement and the converse are both true. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Practice 2-2 1. Two angles have the same measure if and only if they are congruent. 2. The converse, “If ∆n∆ = 17, then n = 17,” is not true. 3. If a whole number is a multiple of 5,then its last digit is either 0 or 5. If a whole number has a last digit of 0 or 5, then it is a multiple of 5. 4. If two lines are perpendicular, then the lines form four right angles. If two lines form four right angles, then the lines are perpendicular. 5. Sample: Other vehicles, such as trucks, fit this description. 6.Sample: Baseball also fits this definition. 7. Sample: Pleasing, smooth, and rigid all are too vague. 8. yes 9. no 10. yes Guided Problem Solving 2-2 1. A good definition is clearly understood, precise, and reversible. 2. &3 and &4, &5 and &6 3. No 4. They are not supplementary. 5. A linear pair has acommon vertex, shares a common side, and is supplementary. 6. yes 7. yes; yes 8. linear pairs: &1 and &2, &3 and &4; not linear pairs: &1 and &3, &1 and &4, &2 and &3, &2 and &4 Practice 2-3 1. Football practice is canceled for Monday. 2. DEF is a right triangle. 3. If two lines are not parallel, then theyintersect at a point. 4. If you vacation at the beach, then you like Florida. 5. Tamika lives in Nebraska. 6. not possible 7. It is not freezing outside. 8. Shannon lives in the smallest state in the United States. L1 All-In-One Answers Version B (continued) Guided Problem Solving 2-3 1. conditional; hypothesis2. Yes 3. Beth will go. 4. Anita, Beth, Aisha, Ramon 5. No; only two students went. 6. Beth, Aisha, Ramon; no—only two went. 7. Aisha, Ramon 8. The answer is reasonable. It is not possible for another pair to go to the concert. 9. Ramon Practice 2-4 1. UT = MN 2. y = 51 3. JL 4. Addition; SubtractionProperty of Equality; Multiplication Property of Equality; Division Property of Equality 5. Substitution 6. Substitution 7. Symmetric Property of Congruence 8. Definition of Complementary Angles; 90, Substitution; 3x, Simplify; 3x, 84, Subtraction Property of Equality; 28, Division Property of Equality GuidedProblem Solving 2-4 1. Angle Addition Postulate 2. Substitution Property of Equality; Simplify; Addition Property of Equality; Division Property of Equality 3. 40 4. yes; yes 5. 13; 13 Practice 2-5 1. 30 2. 15 3. 6 4. m&A = 135; m&B = 45 5. m&A = 10; m&B = 80 6. m&PMO = 55; m&PMQ = 125; m&QMN = 557. m&BWC = m&CWD, m&AWB + m&BWC = 180; m&CWD + m&DWA = 180; m&AWB = m&AWD Guided Problem Solving 2-5 1. 90 2. See graph in Exercise 5 answer. 3. on the positive y-axis 4. Answers may vary. B can be any point on the positive y-axis. Sample: B(0, 3). 5. y B(0,3) A(1,3) X(4,0) xO(0,0) C(3,1) 6. They are adjacent complementary angles. 7. Answers may vary. C can be any point on the line y 5 21 x, x . 0. 3 Sample: C(3, -1). 8. a right angle 9. Yes; their sum corresponds to the right angle formed by the positive x-axis and the positive y-axis. 10. Answers may vary. D can be anypoint on the negative x-axis, sample: D(-4, 0) Geometry 45 Geometry: All-In-One Answers Version B (continued) 2A: Graphic Organizer 2D: Visual Vocabulary Practice 1. Reasoning and Proof 2. Answers may vary. Sample: 1. Law of Detachment 2. hypothesis 3. Distributive Property 4. Reflexive Property5. Law of Syllogism 6. biconditional 7. conclusion 8. good definition 9. Symmetric Property conditional statements; writing biconditionals; converses; and using the Law of Detachment 3. Check students’ work. 2B: Reading Comprehension 2E: Vocabulary Check 1. 42 degrees 2. 38 degrees 3. b Truthvalue: “True” or “false” according to whether the 2C: Reading/Writing Math Symbols statement is true or false, respectively Hypothesis: The part that follows if in an if-then statement. Biconditional: The combination of a conditional statement 1. Segment MN is congruent to segment PQ. 2. If p, then q. 3.The length of MN is equal to the length of PQ. 4. Angle XQV is congruent to angle RDC. 5. If q, then p. 6. The measure of angle XQV is equal to the measure of angle RDC. 7. p if and only if q. 8. a S b 9. AB = MN 10. m/XYZ = m/RPS 11. b S a 12. AB > MN 13. a 4 b 14. /XYZ /RPS 2F: Vocabulary ReviewPuzzle P O S T U L A T 2 V 4 E P R 5 7 C O G C O N S T D I J C C L P A R C O L I N A A J T N E E U T G C E A L T M N E U I G S R C T I O 11 N T D 12 H Y P O L O T H E A L L E L N E E U T I N R N D 8 A O I O 6 T A 9 N 10 A C O R E S I S S I N T Chapter 3 Practice 3-1 1. corresponding angles2. alternate interior angles 3. sameside interior angles 4. 1 and 5, 2 and 6, 3 and 8, 4 and 7 5. 4 and 6, 3 and 5 6. 4 and 5, 3 and 6 7. m1 = 100, alternate interior angles; m2 = 100, corresponding angles or vertical angles 8. m1 = 135, corresponding angles; m2 = 135, vertical angles 9. x = 103; 77°, 103°10. x = 30; 85°, 85° 46 Geometry Guided Problem Solving 3-1 1. The top and bottom sides are parallel, and the left and right sides are parallel. 2. The two diagonals are transversals, and also each side of the parallelogram is a transversal for the two sides adjacent to it. 3. Corresponding angles, interiorand exterior angles are formed. 4. v, w and x; By the Alternate Interior Angles Theorem, v = 42, w = 25 and x = 76. 5. Answers may vary. Possible answer: By the Same-Side Interior Angles Theorem, (w + 42) + (y + 76) = 180. Since w = 25, y = 37. (The two y’s are equal by Theorem 3-1.) 6. w = 25, y = 37,v = 42, x = 76; yes 7. v = 42, w = 35, x = 57, y = 46 All-In-One Answers Version B L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 3 All rights reserved. and its converse; it contains the words “if and only if.” Conclusion: The part of an if-then statement that follows then. Conditional: Anif-then statement. Geometry: All-In-One Answers Version B (continued) Practice 3-2 Practice 3-5 1. l and m, Converse of Same-Side Interior Angles Theorem 2. none 3. BC and AD , Converse of Same-Side Interior Angles Theorem 4. BH and CI, Converse of Corresponding Angles Postulate 5. 43 6. 90 7.38 8. 100 1. x = 120; y = 60 2. n = 5173 3. a = 108; b = 72 4. 109 5. 133 6. 129 7. 30 8. 150 9. 6 10. 5 11. BEDC 12. FAE 13. FAE and BAE 14. ABCDE Guided Problem Solving 3-2 Guided Problem Solving 3-5 1. and m 2. transversals 3. x 4. the angles measuring 19xº and 17xº 5. 180º 6. 17xº 7. 180 - 19x= 17x or 19x+17x = 180 8. x = 5 9. With x = 5, 19x = 95 and 17x = 85. 10. x = 6 1. A theater stage, consisting of a large platform surrounding a smaller platform. The shapes in the bottom part of the figure may represent a ramp for actors to enter and exit. 2. The measures of angles 1 and 2 3. 8; octagon 4.(8 - 2)180 = 1080 degrees 5. 135 6. 45 7. Yes, angle 1 is an obtuse angle and angle 2 is an acute angle. 8. trapezoids 9. 360 All rights reserved. Practice 3-3 1. True. Every avenue will be parallel to Founders Avenue, and therefore every avenue will be perpendicular to Center City Boulevard, andtherefore every avenue will be perpendicular to any street that is parallel to Center City Boulevard. 2. True. The fact that one intersection is perpendicular, plus the fact that every street belongs to one of two groups of parallel streets, is enough to guarantee that all intersections are perpendicular. 3. Notnecessarily true. If there are more than three avenues and more than three boulevards, there will be some blocks bordered by neither Center City Boulevard nor Founders Avenue. 4. a ⊥ e 5. a || e 6. a || e 7. a || e 8. If the number of ⊥ statements is even, then 1 || n. If it is odd, then 1 ⊥ n. Practice 3-6 1. y= 13 x - 7 2. y = -2x + 12 3. y = 45 x - 2 4. y = 4x - 13 5. y y 2x 6. © Pearson Education, Inc., publishing as Pearson Prentice Hall. y 6 4 y 5x 4 2 1. supplementary angles 2. right angle 3. Any one of the following: Postulate 3-1, or Theorem 3-1, 3-2, 3-3 or 3-4 4. 90 5. It is congruent; Postulate 3-1 6. 90 7. a⊥ c 8. It is true for any line parallel to b. 9. Yes. The point is that a 2 4 6 x –6–4–2 –4 –6 y 7. 6 4 2 Practice 3-4 1. 125 2. 143 3. 129 4. 136 5. x = 35; y = 145; z = 25 6. v = 118; w = 37; t = 62 7. 50 8. 88 9. m1 = 33; m2 = 52 10. right scalene 11. obtuse isosceles 12. equiangular equilateral 2 4 6 x –6–4–2 –2 –4 –6 Guided Problem Solving 3-3 transversal cannot be perpendicular to just one of two parallel lines. It has to be perpendicular to both, or else to neither. 6 4 y 4x 1 2 4 6 x –6–4–2 –2 –6 8. y = x + 4 9. y = 12 x - 3 10. y = -21 x - 1 2 Guided Problem Solving 3-4 11. y = -6x + 45 12 y = -11; x = 2 13. y =2; x = 0 1. three 2. 180 3. right triangle 4. z = 90; Because it is given in the figure that BD ⊥ AC. 5. Theorem 3-12, the Triangle Angle-Sum Theorem 6. x = 38 7. y = 36 8. #ABD is a 36-54-90 right triangle. #BCD is a 38-52-90 right triangle. 9. 74 10. #ABC is a 52-54-74 acute triangle. 11. Yes, all three areacute angles, with &ABC visibly larger than &A and &C. 12. &BCD 14. L1 All-In-One Answers Version B y (–2, 0) 6 4 2 –6 –4 –2 2 4 6 x –2 –4 –6 (0, –1) Geometry 47 Geometry: All-In-One Answers Version B Guided Problem Solving 3-7 y 6 (0, 6) 4 2 y K 2 4 6 x –6–4–2 –2 –4 –6 16. 1. (6, 0) H 4 4 6 4 2(4, 0) y B 6. Product = - 85 -1. Sides GH and GK are not perpendicular. 7. #GHK has no pair of perpendicular sides. It is not a right triangle. 8. No 9. &HGK; approximately 80 10. Slope of LM = 72 and slope of LN = - 27 . The product of the slopes is -1, so LM and LN are perpendicular. 2 2 4 2. a right angle3. m1 m2 = -1 4. sides GH and GK 5. Slope of GH = 35 ; slope of GK = 83 Guided Problem Solving 3-6 O A O 4 2 4 6 x –6–4–2 –2 –4 (0, –4) –6 1. x G y 2 x C All rights reserved. 15. (continued) Practice 3-8 1. 2. 2 Q å å AB = 5 ; slope of BC = - 5 . The absolute values of the 2 2 slopes are the same, butone slope is positive and the other is negative. 5. Point-slope form: y - 0 = 5 (x - 0); 2 slope-intercept form: y = 5 x 6. Point-slope form: 2 5 y - 5 = - (x - 2) or y - 0 = - 5 (x - 4); 2 2 å slope-intercept form: y = - 5 x + 10 7. Of line AB : (0, 0); 2 å of line BC : (0, 10) 8. #ABC appears to be an isosceles triangle,which is consistent with a horizontal base and two remiaining sides having slopes of equal magnitude and opposite sign. 9. Slope = 0; y = 0; y-intercept = (0, 0) just as for line å AB (they intersect on the y-axis). Practice 3-7 1. neither; 3 2 13, 3 ? 13 2 -1 2. perpendicular; 2 1 2 2 ? -2 = -1 3. parallel; -3 = -34. perpendicular; T © Pearson Education, Inc., publishing as Pearson Prentice Hall. y 2y 2. m 5 x22 2 x11 3. y - y1 = m(x-x1) 4. Slope of 3. K 4. Sample: b a y = 2 is a horizontal line, x = 0 is a vertical line 5. perpendicular; -1 ? 1 = -1 6. neither; 12 2 -53, 9 9 5 1 2 ? –3 2 -1 7. neither; 2 2 4, 2 ? 4 2 -1 8.parallel; 2 1 1 2 = 2 9. y = 3 x 10. y = 2x - 4 48 Geometry All-In-One Answers Version B L1 Geometry: All-In-One Answers Version B (continued) 7. They appear to be both congruent and parallel. 8. The answers to Step 7 are confirmed. 9. yes; a parallelogram 5. 3A: Graphic Organizer b b b b 1. Paralleland Perpendicular Lines 2. Answers may vary. Sample: properties of parallel lines; finding the measures of angles in triangles; classifying polygons; and graphing lines 3. Check students’ work. 3B: Reading Comprehension 1. 14 spaces 2. 6 spaces 3. 60° 4. corresponding 5. $7000; $480 6. the width of thestalls, 10 ft 7. b 3C: Reading/Writing Math Symbols 6. å å All rights reserved. 1. m # n 2. m/1 + m/2 = 180 3. AB CD 4. m/MNP + m/MNQ = 90 5. /3 /EFD 6. Line 1 is parallel to line 2. 7. The measure of angle ABC is equal to the measure of angle XYZ. 8. Line AB is perpendicular to line DF. 9. Angle ABCand angle ABD are complementary. 10. Angle 2 is a right angle, or the measure of angle 2 is 90°. 11. Sample answer: å å CB GD, m/BAF = m/GFA a c 3D: Visual Vocabulary Practice/High-Use Academic Words Guided Problem Solving 3-8 © Pearson Education, Inc., publishing as Pearson Prentice Hall.1. a line segment of length c 2. Construct a quadrilateral with one pair of parallel sides of length c, and then examine the other pair. 3. The procedure is given on p. 181 of the text. 1. property 2. conclusion 3. describe 4. formula 5. measure 6. approximate 7. compare 8. contradiction 9. pattern 3E:Vocabulary Check m Transversal: A line that intersects two coplanar lines in two points. l Alternate interior angles: Nonadjacent interior angles that lie on opposite sides of the transversal. 4. Adjust the compass to exactly span line segment c, end to end. Then tighten down the compass adjustment asnecessary. 5. m l 6. m l L1 All-In-One Answers Version B Same-side interior angles: Interior angles that lie on the same side of a transversal between two lines. Corresponding angles: Angles that lie on the same side of a transversal between two lines, in corresponding positions. Flow proof: A convincingargument that uses deductive reasoning, in which arrows show the logical connections between the statements. 3F: Vocabulary Review 1. C 2. E 3. D 4. B 5. A 6. F 7. K 8. H 9. L 10. G 11. I 12. J Geometry 49 Geometry: All-In-One Answers Version B 1. m1 = 110; m2 = 120 2. m3 = 90; m4 = 135 3. CA JS, AT SD , CT JD 4. C J, A S, T D 5. Yes; GHJ IHJ by Theorem 4-1 and by the Reflexive Property of . Therefore, GHJ IHJ by the definition of triangles. 6. No; QSR TSV because vertical angles are congruent, and QRS TVS by Theorem 4-1, but none of the sides are necessarily congruent. 7a. Given 7b.Vertical angles are . 7c. Theorem 4-1 7d. Given 7e. Definition of triangles Guided Problem Solving 4-1 1. right triangles 2. m&A = 45, m&B = m&L = 90, and AB = 4 in. 3. #ABC is congruent to #KLM means corresponding sides and angles are congruent. 4. x and t 5. 45 6. m&K = m&M = 45 7. 3x = 45 8. x =15 9. 4 10. 2t = 4 11. t = 2 12. The angle measures indicate that the two triangles are isosceles right triangles. This matches the appearance of the figure. 13. m&M = 60 Practice 4-4 1. BD is a common side, so ADB CDB by SAS, and A C by CPCTC. 2. FH is a common side, so FHE HFG by ASA, and HEFG by CPCTC. 3. QS is a common side, so QTS SRQ by AAS. QST SQR by CPCTC. 4. ZAY and CAB are vertical angles, so ABC AYZ by ASA. ZA AC by CPCTC. 5. JKH and LKM are vertical angles, so HJK MLK by AAS, and JK KL by CPCTC. 6. PR is a common side, so PNR RQP by SSS, and N Qby CPCTC. 7. First, show that ACB and ECD are vertical angles. Then, show ABC EDC by ASA. Last, show A E by CPCTC. Guided Problem Solving 4-4 1. ISOS and SP bisects &ISO. 2. Prove whatever additional facts can be proven about #ISP and #OSP, based on the given information. 3. IS > SO. 4.&ISP &PSO 5. SP 6. #ISP #OSP by Postulate 4-2, the Side-AngleSide(SAS)Postulate 7. It does not matter. The Side-AngleSide Postulate applies whether or not they are collinear. 8. It does follow, because #ISP #OSP and because IP and PQ are corresponding parts. 1. A compass with a fixed settingwas used to draw two circular arcs, both centered at point P but crossing in different locations, which were labeled A and B. The compass was used again, with a wider setting, to draw two intersecting circular arcs, one centered at A and one at B. The point at å which the new arcs intersected was labeledC. Finally, line CP was drawn. 2. Find equal lengths or distances and explain å why CP is perpendicular to . 3. #ACP and #BCP 4. AP = PB and AC = BC. 5. #APC > #BPC, by Postulate 4-1, the Side-Side-Side (SSS) Postulate 6. &APC > &BPC by CPCTC 7. Since &APC > &BPC, m&APC = m&BPC andm&APC + m&BPC = 180, it follows that m&APC = m&BPC = 90. 8. from the definition of perpendicular and the fact that m&APC = m&BPC = 90 9. The distances do not matter, so long as AP = BP and AC = BC. That is what is required in order that #APC > #BPC. 10. Draw a line, and use the constructiontechnique of the problem to construct a second line perpendicular to the first. Then do the same thing again to construct a third line perpendicular to the second line. The first and third lines will be parallel, by Theorem 3-10. Practice 4-3 Practice 4-5 Practice 4-2 1. ADB CDB by SAS 2. not possible 3. TUSXWV by SSS 4. not possible 5. DEC GHF by SAS 6. PRN PRQ by SSS 7. C 8. AB and BC 9. A and B 10. AC 11a. Given 11b. Reflexive Property of Congruence 11c. SAS Postulate Guided Problem Solving 4-2 1. not possible 2. ASA Postulate 3. AAS Theorem 4. ASA Postulate 5. not possible 6. AASTheorem 7. Statements Reasons 1. K M, KL ML 1. Given 2. JLK PLM 2. Vertical s are . 3. JKL PML 3. ASA Postulate 8. BC EF 9. KHJ HKG or KJH HGK Guided Problem Solving 4-3 1. Corresponding angles and alternate interior angles. 2. &EAB and &DBC. 3. &EBA and &DCB. 4. &EAB 50 Geometry Allrights reserved. Practice 4-1 and &DBC 5. &EAB > &DBC, AE > BD, and &E > &D. 6. #AEB > #BDC by Postulate 4-3, the AngleSide-Angle (ASA) Postulate 7. Yes; Theorem 4-2, the Angle-Angle-Side (AAS) Theorem; &EBA > &DCB. 8. No, because now there is no way to demonstrate a second pair ofcongruent sides, nor a second pair of congruent angles. 1. x = 35; y = 35 2. x = 80; y = 90 3. t = 150 4. x = 55; y = 70; z = 125 5. x = 6 6. z = 120 7. AD ; D F 8. KJ ; KIJ KJI 9. BA ; ABJ AJB 10. 130 11. 130 12. x = 70; y = 55 Guided Problem Solving 4-5 1. One angle is obtuse. The other two angles areacute and congruent. 2. Highlight an obtuse isosceles angle and find its angle measures, then find all the other angle measures represented in the figure. All-In-One Answers Version B L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Chapter 4 (continued) Geometry: All-In-One AnswersVersion B (continued) 3. Possible answer: 8. hypotenuse shorter leg = 2; yes, this matches the figure. 9. The solution remains the same: x = 3 and y = 2. The reason is that one is still solving the same two equations, x = y + 1 and x + 3 = 3y. Practice 4-7 4. 60 5. 30, because the measure of each baseangle is half the measure of an angle of the equilateral triangle. 6. 120° 1. ZWX YXW; SAS 2. LNP LMO; SAS 3. ADF BGE; SAS 4. F common side: FG G G F because the sum of the angles of the highlighted triangle must equal 180°. 7. The other measures are 90° and 150°. Examples: E All rightsreserved. 5. common angle: L L L N K J M 6. Sample: 8. Yes; 3 120 = 360. 9. Answers may vary. Practice 4-6 1. Statements 1. AB # BC , ED # FE 2. B, E are right s. © Pearson Education, Inc., publishing as Pearson Prentice Hall. H 3. AC FD , AB ED 4. ABC DEF Reasons 1. Given 2. Perpendicular linesform right s. 3. Given 4. HL Theorem MJN and MJK are right s. 2. Perpendicular lines form right s. MN MK MJN MJK Given HL Theorem MJ MJ Reflexive Property of 3. RS VW 4. none 5. mC and mF = 90 6. ST UV or SV UT 7. mA and mX = 90 8. GI # JH Guided Problem Solving 4-6 1. Two congruentright triangles. Each one has a leg and a hypotenuse labeled with a variable expression. 2. the values of x and y for which the triangles are congruent by HL 3. The two shorter legs are congruent. 4. x = y + 1 5. The hypotenuses are congruent. 6. x + 3 = 3y 7. x = 3; y = 2 L1 All-In-One Answers Version BStatements 1. AX AY 2. CX # AB, BY # AC 3. mCXA and mBYA = 90 4. A A 5. BYA CXA Reasons 1. Given 2. Given 3. Perpendicular lines form right s. 4. Reflexive Property of 5. ASA Postulate Guided Problem Solving 4-7 1. The figure, a list of parallel and perpendicular pairs of sides, and one knownangle measure, namely m&A = 56 2. nine 3. They are congruent and have equal measures. 4. m&A = m&1 = m&2 = 56 5. m&4 = 90 6. m&3 = 34 7. m&DCE = 56 8. m&5 = 22 9. m&FCG = 90 10. m&6 = 34 11. m&7 = 34, m&8 = 68, and m&9 = 112 12. m&9 = 56 + 56 = 112 13. m&FIC = 180 -(m&2 + m&3)= 90; m&DHC = m&4 = 90; m&FJC = 180 - m&9 = 68; m&BIG = m&FIC = 90 4A: Graphic Organizer 1. Congruent Triangles 2. Answers may vary. Sample: congruent figures; triangle congruence by SSS, SAS, ASA, and AAS; proving parts of triangles congruent; the Isosceles Triangle Theorem 3. Checkstudents’ work. 4B: Reading Comprehension 1. Yes. Using the Isosceles Triangle Theorem, /W /Y. It is given that WX YX and WU YV. Therefore nWUX nYVX by SAS. 2. There is not enough information. You need to know if AC EC, if /A /E, or if /B /D. 3. a Geometry 51 (continued) 4C: Reading/WritingMath Symbols Guided Problem Solving 5-2 1. Angle-Angle-Side 2. triangle XYZ 3. angle PQR 4. line segment BD 5. line ST 6. ray WX 7. hypotenuseleg 8. line 3 9. angle 6 10. Angle-Side-Angle 1. y 4 1. theorem 2. congruent polygons 3. base angle of an isosceles triangle 4. CPCTC 5. postulate 6. vertexangle of an isosceles triangle 7. corollary 8. base of an isosceles triangle 9. legs of an isosceles triangle O corresponding sides congruent and corresponding angles congruent. CPCTC: An abbreviation for “corresponding parts of congruent triangles are congruent.” 4F: Vocabulary Review Puzzle 1.postulate 2. hypotenuse 3. angle 4. vertex 5. side 6. leg 7. perpendicular 8. polygon 9. supplementary 10. parallel 11. corresponding Chapter 5 Practice 5-1 1a. 8 cm 1b. 16 cm 1c. 14 cm 2a. 9.5 cm 2b. 17.5 cm 2c. 14.5 cm 3. 17 4. 7 5. 42 6. 16.5 7a. 18 7b. 61 8. PR 6 YZ, PQ 6 XZ, XY 6 RQ GuidedProblem Solving 5-1 1. 30 units 2. The three sides of the large triangle are each bisected by intersections with the two line segments lying in the interior of the large triangle. 3. the value of x 4. They are called midsegments. 5. They are parallel, and the side labeled 30 is half the length of the side labeled x.

6. x = 60 7. Yes; the side labeled x appears to be about twice as long as the side labeled 30. 8. No. Those lengths are not fixed by the given information. (The triangle could be vertically stretched or shrunk without changing the lengths of the labeled sides.) All one can say is that the midsegment is half aslong as the side it is parallel to. Practice 5-2 1. WY is the perpendicular bisector of XZ. 2. 4 3. 9 4. right triangle 5. 5 6. 17 7. )isosceles triangle 8. 3.5 9. 21 10. right triangle 11. JP is the bisector of LJN. 12. 9 13. 45 14. 14 15. right isosceles triangle 52 Geometry 4 C B x 2. See answer to Step 1, above. 3.See answer to Step 1, above. 4. Plot a point and explain why it lies on the bisector of the angle at the origin. 5. line : y 5 23 x 1 25 ; line m: 4 2 x = 10 6. C(10, 5) 7. CA = CB = 5; yes 8. Theorem 5-5, the Converse of the Angle Bisector Theorem 9. m&AOC = m&BOC < 27 10. Draw , m, and C, then drawOC. Since OA = OB = 10, it follows that #OAC > #OBC, by HL. Then CA > CB and &AOC > &BOC by CPCTC. Practice 5-3 1. (-2, 2) 2. (4, 0) 3. altitude 4. median 5. perpendicular bisector 6. angle bisector 7a. (2, 0) 7b. (-2, -2) 8a. ( 0, 0) 8b. (3, -4) Guided Problem Solving 5-3 1. the figure and a proof withsome parts left blank 2. Fill in the blanks. 3. AB 4. Theorem 5-2, the Perpendicular Bisector Theorem 5. BC; XC 6. the Transitive Property of Equality 7. Perpendicular Bisector. (This converse is Theorem 5-3.) 8. The point of the proof is to demonstrate that n runs through point X. It would not beappropriate to show that fact as already given in the figure. 9. Nothing essential would change. Point X would lie outside #ABC å (below BC ), but the proof woud run just the same. Practice 5-4 1. I and III 2. I and II 3. The angle measure is not 65. 4. Tina does not have her driver’s license. 5. The figuredoes not have eight sides. 6. ABC is congruent to XYZ. 7a. If you do not live in Toronto, then you do not live in Canada; false. 7b. If you do not live in Canada, then you do not live in Toronto; true. 8. Assume that mA 2 mB. 9. Assume that LM does not intersect NO. 10. Assume that it is not sunny outside.11. Assume that mA 90. This means that mA + mC 180. This, in turn, means that the sum of the angles of ABC exceeds 180, which contradicts the Triangle Angle-Sum Theorem. So the assumption that mA 90 must be incorrect. Therefore, mA 90. Guided Problem Solving 5-4 1. Ice is forming on thesidewalk in front of Toni’s house. 2. Use indirect reasoning to show that the temperature of the sidewalk surface must be 32°F or lower. 3. The temperature All-In-One Answers Version B L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Angle: Formed by two rays with the same endpoint.Congruent angles: Angles that have the same measure. Congruent segments: Segments that have the same length. Corresponding polygons: Polygons that have m A 4D: Visual Vocabulary Practice 4E: Vocabulary Check < All rights reserved. Geometry: All-In-One Answers Version B Geometry: All-In-One Answers Version B (continued) All rights reserved. of the sidewalk in front of Toni’s house is greater than 32°F. 4. Water is liquid (ice does not form) above 32°F. 5. There is no ice forming on the sidewalk in front of Toni’s house. 6. The result from step 5 contradicts the information identified as given instep 1. 7. The temperature of the sidewalk in front of Toni’s house is less than or equal to 32°F. 8. If the temperature is above 32°F, water remains liquid. This is reliably true. Converse: If water remains liquid, the temperature is above 32°F. This is not reliably true. Adding salt will cause water to remainliquid even below 32°F. 9. Suppose two people are each the world’s tallest person. Call them person A and person B. Then person A would be taller than everyone else, including B, but by the same token B would be taller than A. It is a contradiction for two people each to be taller than the other. So it isimpossible for two people each to be the World’s Tallest Person. Practice 5-5 1. M, N 2. C, D 3. R, P 4. A, T 5. yes; 4 + 7 8, 7 + 8 4, 8 + 4 7 6. no; 6 + 10 17 7. yes; 4 + 4 4 8. yes; 11 + 12 13, 12 + 13 11, 13 + 11 12 9. no; 18 + 20 40 10. no; 1.2 + 2.6 4.9 11. BO, BL, LO 12. RS, ST, RT 13. D, S, A 14. N, S,J 15. 3 x 11 16. 8 x 26 17. 0 x 10 18. 9 x 31 Guided Problem Solving 5-5 1. 5C: Reading/Writing Math Symbols 1. L 2. F 3. O 4. G 5. A 6. I 7. M 8. H 9. E 10. K 11. B 12. D 13. N 14. J 15. C 5D: Visual Vocabulary Practice 1. median 2. negation 3. circumcenter 4. contrapositive 5. centroid 6. equivalentstatements 7. incenter 8. inverse 9. altitude 5E: Vocabulary Check Midpoint: A point that divides a line segment into two congruent segments. Midsegment of a triangle: The segment that joins the midpoints of two sides of a triangle. Proof: A convincing argument that uses deductive reasoning. Coordinateproof: A proof in which a figure is drawn on a coordinate plane and the formulas for slope, midpoint, and distance are used to prove properties of the figure. Distance from a point to a line: The length of the perpendicular segment from the point to the line. 5F: Vocabulary Review 1. altitude 2. line 3. median4. negation 5. contrapositive 6. incenter 7. orthocenter 8. slope-intercept 9. exterior 10. obtuse 11. alternate interior 12. centroid 13. equivalent 14. right 15. parallel © Pearson Education, Inc., publishing as Pearson Prentice Hall. Chapter 6 Practice 6-1 2. The side opposite the larger included angle isgreater than the side opposite the smaller included angle. 3. The angle opposite the larger side is greater than the angle opposite the smaller side 4. The opposite sides each have a length of nearly the sum of the other two side lengths. 5. The opposite sides are the same length. They are correspondingparts of triangles that are congruent by SAS. 5A: Graphic Organizer 1. Relationships Within Triangles 2. Answers may vary. Sample: midsegments of triangles; bisectors in triangles; concurrent lines, medians, and altitudes; and inverses, contrapositives, and indirect reasoning 3. Check students’ work. 1.parallelogram 2. rectangle 3. quadrilateral 4. kite, quadrilateral 5. trapezoid, isosceles trapezoid, quadrilateral 6. square, rectangle, parallelogram, rhombus, quadrilateral 7. x = 7; AB = BD = DC = CA = 11 8. f = 5; g = 11; FG = GH = HI = IF = 17 9. parallelogram 10. kite Guided Problem Solving 6-1 1. alabeled figure, which shows an isosceles trapezoid 2. The nonparallel sides are congruent. 3. the measures of the angles and the lengths of the sides 4. m&G = c 5. c + (4c - 20) = 180 6. 40 7. m&D = m&G = 40, m&E = m&F = 140 8. a - 4 = 11 9. 15 10. DE = FG = 11, EF = 15, DG = 32 11. 40 + 40 + 140 +140 = 360 = (4 - 2)180 12. m&D = m&G = 39, m&E = m&F = 141 Practice 6-2 5B: Reading Comprehension 1. The width of the tar pit is 10 meters. 2. b L1 All-In-One Answers Version B 1. 15 2. 32 3. 7 4. 12 5. 9 6. 8 7. 100 8. 40; 140; 40 9. 113; 45; 22 10. 115; 15; 50 11. 55; 105; 55 12. 32; 98; 50 13. 1614. 35 15. 28 Geometry 53 Geometry: All-In-One Answers Version B x know AB > AD, but it was necessary to know m&B = 90. The key fact, which enables the use of Theorem 6-11 in addition to Theorem 6-3, is that ABCD is a rectangle. It does not matter whether all four sides are congruent. 9. 40 4. themeasures of the angles 5. The angles are Practice 6-5 supplementary angles, because they are consecutive. 6. x + 9x = 180 7. 18 and 162 8. No; the lengths of the sides are irrelevant in this problem. 9. 30 and 150 1. 118; 62 2. 59; 121 3. 96; 84 4. 101; 79 5. x = 4 6. x = 1 7. 105.5; 105.5 8. 118; 118 9. 90;63; 63 10. 107; 107 11. x = 8 12. x = 7 Practice 6-3 Guided Problem Solving 6-5 1. no 2. yes 3. yes 4. yes 5. x = 2; y = 3 6. x = 64; y = 10 7. x = 8; the figure is a because both pairs of opposite sides are congruent. 8. x = 25; the figure is a because the congruent opposite sides are 6 by the Converse of theAlternate Interior Angles Theorem. 9. No; the congruent opposite sides do not have to be 6. 10. No; the figure could be a trapezoid. 11. Yes; both pairs of opposite sides are congruent. 12. Yes; both pairs of opposite sides are 6 by the converse of the Alternate Interior Angles Theorem. 13. No; only onepair of opposite angles is congruent. 14. Yes; one pair of opposite sides is both congruent and 6. 1. Isosceles trapezoid ABCD with AB > DC 2. &B > &C and &BAD > &D 3. AB > DC is Given. DC > AE because opposite sides of a parallelogram are congruent (Theorem 6-1). AB > AE is from the TransitiveProperty of Congruency. 4. Isosceles; >; because base angles of an isosceles triangle are congruent. 5. &1 > &C because corresponding angles on a transversal of two parallel lines are congruent. 6. &B > &C by the Transitive Property of Congruency. 7. &BAD is a same-side interior angle with &B, and&D is a same-side interior angle with &C. 8. This is not a problem, because for AD > BC there is a similar proof with a line segment drawn from B to a point E lying on AD. 9. The two drawn segments can be shown to be congruent, and then one has two congruent right triangles by the HL Theorem. &B >&C follows by CPCTC and &BAD > &D because they are supplements of congruent angles. Guided Problem Solving 6-3 1. a labeled figure, which shows a quadrilateral that appears to be a parallelogram 2. The consecutive angles are supplementary. 3. find values for x and y which make the quadrilaterala parallelogram 4. m&A + m&D = 180, so that &A and &D meet the requirements for same-side interior angles on the transversal of two parallel lines (Theorems 3-2 and 3-6). 5. &B > &D, by Theorem 6-2. 6. 3x + 10 + 5y = 180; 8x + 5 = 5y 7. x = 15, y = 25 8. m&A = m&C = 55 and m&B = m&D = 125,which matches the appearance of the figure. 9. (3x + 10) + (8x + 5) = 180; yes Practice 6-4 1a. rhombus 1b. 72; 54; 54; 72 2a. rectangle 2b. 37; 53; 106; 74 3a. rectangle 3b. 60; 30; 60; 30 4a. rhombus 4b. 22; 68; 68; 90 5. Possible; opposite angles are congruent in a parallelogram. 6. Impossible; if thediagonals are perpendicular, then the parallelogram should be a rhombus, but the sides are not of equal length. 7. x = 7; HJ = 7; IK = 7 8. x = 6; HJ = 25; IK = 25 9a. 90; 90; 29; 29 9b. 288 cm2 10a. 38; 90; 90; 38 10b. 260 m2 Guided Problem Solving 6-4 1. A labeled figure, which shows a parallelogram.One angle is a right angle, and two adjacent sides are congruent. Algebraic expressions are given for the lengths of three line segments. 2. diagonals 3. Find the values of x and y. 4. It is a square. Theorem 6-1 and the fact that AB > AD imply that all four sides are congruent. Theorems 3-11 and 6-2 plusthe fact that m&B = 90 imply that all four angles are right angles. 54 Geometry Practice 6-6 1. (1.5a, 2b); a 2. (0.5a, 0); a 3. (0.5a, b); " a2 1 4b2 2b 4. 0 5. 1 6. -12 7. 2b 3a 8. -3a 9. E(a, 3b); I(4a, 0) 10. D(4a, b); I(3a, 0) 11. (-4a, b) 12. (-b, 0) Guided Problem Solving 6-6 1. a rhombus with coordinates givenfor two vertices 2. Arhombus is a parallelogram with four congruent sides. 3. the coordinates of the other two vertices 4. They are the diagonals. 5. They bisect each other. 6. W(-2r, 0), Z(0, -2t) 7. No; neither Theorem 6-3 nor any other theorem or result would apply. 8. Slope of WX = 0 and slope of YZ isundefined. This confirms Theorem 6-10, which says that the diagonals of a rhombus are perpendicular. Practice 6-7 p p2 p q 1b. y = mx + b; q = q (p) + b; b = q - q ; p p p2 y = q x + q - q 1c. x = r + p 1d. y = q (r + p) + 2 2 2 rp rp p p p q - q ; y = q + q + q - q ; y = q + q; intersection rp at (r + p, q + q) 1e. qr1f. (r, q) 1g. y = mx + b; 2 2 q = qr (r) + b; b = q - rq ; y = rqx + q - rq 2 2 2 rp 1h. y = qr (r + p) + q - rq ; y = rq + q + q - rq ; 1a. All-In-One Answers Version B L1 All rights reserved. 1. the ratio of two different angle measures in a parallelogram 2. The consecutive angles are supplementary. 3. 9x 5.congruent; bisect 6. 4x - y + 1 = (2x - 1) + (3y + 5); 2x - 1 = 3y + 5 7. x = 7 1 ; y = 3 8. It was not necessary to 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Guided Problem Solving 6-2 (continued) Geometry: All-In-One Answers Version B (continued) rp rp Guided Problem Solving 6-7the given information, only one side and one angle of the two triangles can be proven to be congruent. Another side or angle is needed. If it were given that QUSV is a parallelogram, then the proof could be made. 3. All four sides are congruent. 4. Yes. Since EG > EG by the Reflexive Property, nEFG >nEHG by SSS. 5. b 1. kite DEFG with DE = EF with the midpoint of each side identified 2. A kite is a quadrilateral with two pairs of 6C: Reading/Writing Math Symbols y = q + q; intersection at (r + p, q + q) rp 1i. (r + p, q + q) 2a. (-2a, 0) 2b. (-a, b) All rights reserved. b 2c. (-3a 2 , 2 ) 2d. b a adjacent sidescongruent and no opposite sides congruent. 3. The midpoints are the vertices of a rectangle. 4. D(-2b, 2c), G(0, 0) 5. L(b, a + c), M(b, c), N(-b, c), K(-b, a + c) 6. Slope of KL = slope of NM = 0, slopes of KN and LM are undefined 7. Opposite sides are parallel; it is a rectangle. 8. Adjacent sides areperpendicular. 9. right angles 10. Answers will vary. Example: a = 3, b = 2, c = 2 yields the points D(-4, 4), E(0, 6), F(4, 4), G(0, 0) with midpoints at (-2, 2), (-2, 5), (2, 5), and (2, 2). Connecting these midpoints forms a rectangle. 11. Construct DF and EG. Slope of DF = 0, so DF is horizontal. Slope of EG isundefined, so EG is vertical. 1. AH, CH, or BH 2. DG, FG, or EG 3. G 4. DE 5. rhombus 6. rectangle 7. square 8. isosceles trapezoid 6D: Visual Vocabulary Practice/High-Use Adademic Words 1. solve 2. deduce 3. equivalent 4. indirect 5. equal 6. analysis 7. identify 8. convert 9. common 6E: VisualVocabulary Check Consecutive angles: Angles of a polygon that share a common side. Kite: A quadrilateral with two pairs of congruent adjacent sides and no opposite sides congruent. Parallelogram: A quadrilateral with two pairs of parallel sides. Rhombus: A parallelogram with four congruent sides.Trapezoid: A quadrilateral with exactly one pair of parallel sides. 6A: Graphic Organizer 1. Quadrilaterals 2. Answers may vary. Sample: classifying quadrilaterals; properties of parallelograms; proving that a quadrilateral is a parallelogram; and special parallelograms 3. Check students’ work. 6B: ReadingComprehension © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. QT > SR, QR > ST, QT 6 RS, QR 6 TV 2. No, it cannot be proven that nQTV > nSRU because with 6F: Vocabulary Review Puzzle 1 2 P A R A L 4 T L 5 R E L A P O E G Z R A O I H 7 O M 8 B U S 9 C 3 I M N I C D E SN E N T 6 R O G E R A M R C S E U E N M A T C G All-In-One Answers Version B I B Q U A R N L1 C D E 10 H Y P O T E N L T E E S R U Geometry S E 55 Geometry: All-In-One Answers Version B (continued) Guided Problem Solving 7-4 Chapter 7 1. #ABC, #ACD, #BCD 2. 1 3. 1; 1 4. 1 5. 1 6. 2; 17. 2 8. "2 9. "2 10. yes 11. no (This would 1. 1 : 278 2. 18 ft by 10 ft 3. 18 ft by 16 ft 4. 10 ft by 3 ft 5. true 6. false 7. true 8. false 9. true 10. true 11. 12 12. 12 13. 33 14. 48 15. 6 16. 52 17. 14 : 5 7 18. 12 : 7 19. 38 20. 13 Guided Problem Solving 7-1 42 1 or 1. ratios 2. 42,000,000 3. the denominator1,000,000 x 5 1 4. 29,000 5. Cross-Product Property 6. 0.029 1,000,000 7. 0.348 8. yes 9. 21.912 require the Pythagorean Theorem.) Practice 7-5 1. BE 2. BC 3. JD 4. BE 5. 16 3 6. 4 7. x = y = 4 8. 15 9. x = 6; y = 6 10. 2 11. 10 4 25 9; Guided Problem Solving 7-5 1. parallel 2. CE; BD 3. 6; 15 4. 90 5. yes6. yes 7. The sides would not be parallel. 8. They are similar triangles. 1. ABC XYZ, with similarity ratio 2 : 1 2. Not similar; corresponding sides are not proportional. 3. Not similar; corresponding angles are not congruent. 4. ABC KMN, with similarity ratio 4 : 7 5. I 6. O 7. NO 8. LO 9. 3.96 ft 10. 3.75 cm 11.32 12. 53 13. 7 21 14. 4 12 15. 37 Guided Problem Solving 7-2 1. equal 2. 6.14 3. 2.61; 6.14 4. 19.3662; 19.2182 5. no 2.61 6. no 7. 2.3706; 2.3525 8. Since the quotients are not equal, the ratios are not equal, and the bills are not similar rectangles. 9. 4.045 Practice 7-3 1. AXB RXQ because verticalangles are . A R (Given). Therefore AXB RXQ by the AA 3 XM PX Postulate. 2. Because MP LW = WA = AL = 4 , MPX LWA by the SSS Theorem. 3. QMP AMB QM 2 because vertical s are . Then, because AM = PM BM = 1 , QMP AMB by the SAS Theorem. 4. Because BX AX = BX and CX = RX, AXCX = RX . AXB CXR because vertical angles are . Therefore AXB CXR by 20 48 the SAS Theorem. 5. 15 2 6. 7 7. 3 8. 36 9. 33 ft All rights reserved. 7A: Graphic Organizer Practice 7-2 1. Similarity 2. Answers may vary. Sample: ratios and proportions; similar polygons; proving triangles similar; andsimilarity in right triangles 3. Check students’ work. 7B: Reading Comprehension 1. DHB ACB 2. AA similarity postulate The triangles have two similar angles. 3. 1 : 2 4. 1 : 2 5 HB 6. 250 ft 7. a 5. DB AB CB 7C: Reading/Writing Math Symbols 1. no 2. no 3. yes 4. no 5. yes 6. no 7. yes, AAS 8. yes,Hypotenuse-Leg Theorem 9. yes, SAS or ASA or AAS 10. not possible 7D: Visual Vocabulary Practice 1. Angle-Angle Similarity Postulate 2. golden ratio 3. Side-Side-Side Similarity Theorem 4. geometric mean 5. scale 6. Cross-Product Property 7. golden rectangle 8. simplest radical form 9. Side-Angle-Side Similarity Theorem 7E: Vocabulary Check 1. no; N/A 2. yes; WT, RS 3. It is a trapezoid. 4. They are congruent. 5. They are parallel. 6. They are congruent. 7. #RSZ and #TWZ 8. AA~ or Angle-Angle Similarity Postulate 9. No; there is only one pair of congruent angles. 10. yes; parallelogram,rhombus, rectangle, and square Similarity ratio: The ratio of lengths of corresponding sides of similar polygons. Cross-Product Property: The product of the extremes of a proportion is equal to the product of the means. Ratio: A comparison of two quantities by division. Golden rectangle: A rectangle thatcan be divided into a square and a rectangle that is similar to the original rectangle. Scale: The ratio of any length in a scale drawing to the corresponding actual length. Practice 7-4 7F: Vocabulary Review Guided Problem Solving 7-3 1. 16 2. 8 3. 10"2 4. 6"5 5. h 6. y 7. a 8. c 9. 92 10. x = 6; y = 6"3 11. x =4 "5 ; y = "55 12. 2"15 in. 56 Geometry 1. L 2. E 3. I 4. A 5. J 6. K 7. C 8. D 9. G 10. O 11. N 12. M 13. H 14. B 15. F All-In-One Answers Version B L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Practice 7-1 Geometry: All-In-One Answers Version B (continued) Practice 8-5 Chapter 8Practice 8-1 1. " 51 2. 2 " 65 3. 2 " 21 4. 18 " 2 5. 46 in. 6. 78 ft 7. 279 cm 8. 19 m 9. acute 10. obtuse 11. right 1a. angle of depression from the plane to the person 1b. angle of elevation from the person to the plane 1c. angle of depression from the person to the sailboat 1d. angle of elevation from thesailboat to the person 2. 116.6 ft 3. 84.8 ft 4. 46.7 ft 5. 31.2 yd 6a. Guided Problem Solving 8-1 1. the sum of the lengths of the sides 2. Pythagorean Theorem 3. 7 cm 4. 4 cm 3 cm 5. c2 = 42 + 32 6. 5 7. 12 cm 8. perimeter of rectangle = 14 cm; yes 9. Answers will vary; 35 30 ft All rights reserved.example: Draw a 4 cm 3 cm grid, copy the given figure, measure the lengths with a ruler, add them together. 10. 20 cm 6b. 26 ft Practice 8-2 1. x = 2; y = " 3 2. 8 " 2 3. 14 " 2 4. 2 5. x = 15; y = 15 " 3 6. 3 " 2 7. 42 cm 8. 10.4 ft, 12 ft 9. a = 4; b = 3 10. p = 4 " 3; q = 4 " 3; r = 8; s = 4 " 6 Guided ProblemSolving 8-2 1. 30°-60°-90° triangle 2. l 3. h 4. Í 3 5. 24 or 8Í 3 !3 6. 2 7. 48 or 16Í 3 8. 28 ft 9. 0.28 min 10. yes !3 11. 34 ft 1. &e = 1, &d = &4 2. congruent 3. m&e = m&d 4. 7x - 5 = 4(x + 7) 5. 11 6. 72 7. 72 8. yes 9. 44, 44 Practice 8-6 1. 46.0, 46.0 2. 89.2, -80.3 3. 38.6 mi/h; 31.2º north of east 4. 134.5 m;42.0º south of west 5. 55º north of east 6. 33º west of north 7a. 1, 5 1. tan E = 34 ; tan F = 34 2. tan E = 52; tan F = 52 3. 12.4 4. 31.0° 5. 7.1 6. 6.4 7. 26.6 8. 71.6 9. 39 10. 72 11. 39 12. 54 A B 6 4 2 4 6x 642 2 4 6 Guided Problem Solving 8-3 1. y 7b. Practice 8-3 © Pearson Education, Inc., publishing asPearson Prentice Hall. Guided Problem Solving 8-5 8a. 1, -1 8b. 20 cm y 6 4 80 cm 642 2 4 6 2. 180 3. m&A = 2m&X 4. 90 5. base: 40 cm, height: 10 cm 6. 4 7. 4 8. 76 9. 152 10. 28 11. yes 12. 46 Practice 8-4 1. sin P = 2"710 ; cos P = 3 7 = 65 3 5 8 = 17 2. sin P = 45; cos P = 15 3. sin P = "611 ; cos P4. sin P = 17 ; cos P 5. 64 6. 11.0 7. 7.0 8. 7.8 9. 53 10. 6.6 11. 11.0 12. 11.5 4 6x 9. Sample: N W 48 E Guided Problem Solving 8-4 1. The sides are parallel. 2. sine 3. sin 30° 5 w6 4. 3.0 5. yes 6. cosine 7. cos x° 5 34 8. 41 9. Answers may vary. Sample: cos 60° 36; sin 49° 34 10. 5.2, 2.6 L1 All-In-OneAnswers Version B S Geometry 57 Geometry: All-In-One Answers Version B (continued) Guided Problem Solving 8-6 8D: Visual Vocabulary Practice x ; y 1. Check students’ work. 2. 100 100 1. 30°-60°-90° triangle 2. inverse of tangent 3. congruent sides 4. tangent 5. Pythagorean Theorem 6. hypotenuse7. 45°-45°-90° triangle 8. Pythagorean triple 9. obtuse 3. 100 cos 30°; 100 sin 30° 4. 86.6; 50 5. 86.6, 50 6. 86.6, -50 7. 173.2, 0 8. 173; due east 9. yes 10. 100; due east triangle 8A: Graphic Organizer 8E: Vocabulary Check 1. Right Triangles and Trigonometry 2. Answers may vary. Obtuse triangle: Atriangle with one angle whose measure is Sample: the Pythagorean Theorem; special right triangles; the tangent ratio; sine and cosine ratios; angles of elevation and depression; vectors 3. Check students’ work. Isosceles triangle: A triangle that has at least two congruent between 90 and 180. sides.Hypotenuse: The side opposite the right angle in a right Right triangle: A triangle that contains one right angle. Pythagorean triple: A set of three nonzero whole numbers 1. A 2. J 3. B 4. J 5. B 6. B 7. b a, b, and c that satisfy the equation a2 + b2 = c2. 8C: Reading/Writing Math Symbols 1. F 2. G 3. D 4. A5. C 6. B 7. H 8. E 5 10. #ABC , #XYZ 11. m&A < 52° 9. sin21 A 5 12 7 12. tan Z 5 24 58 J A T N A T L U S E R Y G D H P G H S N L P U R F S I M E E M Q O I I Y H G E A K U R R E L O F L S B T N T N L O N Q G M A M M D M T P B K V I X E X X E C S Q E A L V Y D W M E T D O T S B U N T PX L T Z I T O R N I F A D R Y R L P A H S R P S Z S E O E E E J I I M T A H Y M C O N E D C L M Z C R D E G H E D P I L U T I N E B M J C V O R F P S U T A B A F N V E C T O R Y O O R I N I V G N T M A D D K E B C O N G R U E N T G H N T B R A A N G L E O F D E P R E S S I O N R A E UA R A X L L Y T N E J O B M C O O V Z L P R O P O R T I O N Geometry All-In-One Answers Version B © Pearson Education, Inc., publishing as Pearson Prentice Hall. 8F: Vocabulary Review Puzzle All rights reserved. triangle. 8B: Reading Comprehension L1 Geometry: All-In-One Answers Version B(continued) 8. Chapter 9 4 A 2 Practice 9-1 1. No; the triangles are not the same size. 2. Yes; the ovals are the same shape and size. 3a. C and F 3b. CD and CD, DE and DE, EF and EF, CF and CF 4. (x, y) S (x - 2, y - 4) 5. (x, y) S (x + 4, y - 2) 6. (x, y) S (x + 2, y + 2) 7. W(-2, 2), X(-1, 4), Y(3, 3), Z(2, 1)8. J(-5, 0), K(-3, 4), L(-3, -2) 9. (x, y) S (x + 13, y - 13) 10. (x, y) S (x,y) (x + 3, y + 3) 11a. P(-3, -1) 11b. P(0, 8), N(-5, 2), Q(2, 3) 6 4 2 2 A 9. All rights reserved. 1. the four vertices of a preimage and one of the vertices of the image 2. Graph the image and preimage. 3. C(4, 2) and C(0, 0) 4. x = 4, y = 2, x +a = 0, y + b = 0 5. a = -4; b = -2; (x, y) → (x - 4, y - 2) 6. A(-1, 4), B(1, 3), 6 x 2 4 6 x 4 B T T A 4 2 y 4 6 4 2 Guided Problem Solving 9-1 T 4 B B A 10. (-6, 4) D(-2, 1) 7. y Guided Problem Solving 9-2 A B A B D 1. a point at the origin, and two reflection lines 2. A reflection is an isometry in which a figureand its image have opposite orientations. 3. the image after two successive reflections 4. C D 2 O C 2 x y y=3 4 x 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. y T B O(0,0) 4 8. yes 9. (x, y) → (x + 1, y - 3), A’(4, 3), B’(6, 2), D’(3, 0) Practice 9-2 1. (-3, -2) 2. (-2, -3) 3. (-1, -4) 4. (4, -2)5. (4, -1) 6. (3, -4) y 7a. 5. y=3 y O'(0,6) 4 Z x 2 O(0,0) Y 6 4 2 2 4 W 4 6x X 6. y=3 7b. 4 y O'(0,6) y 4 X 6 4 Y 2 x W Z 4 O(0,0) 6x 4 L1 All-In-One Answers Version B 4 O"(0,6) 7. (0, - 6) 8. Yes. The x-coordinate remains 0 throughout. 9. O(0, 0) and O(0, 6) Geometry 59 Geometry: All-In-One AnswersVersion B Practice 9-3 1. I 2. I 3. I 4. GH 5. G 6. ST 7. (continued) 5. slope of OB 5 252 ; slope of OC 5 52 ; slope of OD 5 252 ; the slopes of perpendicular line segments are negative reciprocals. 6. square 7. yes 8. B(2, 7), C(7, -2), D(-2, -7) P Practice 9-4 1. The helmet has reflectional symmetry. 2. Theteapot has reflectional symmetry. 3. The hat has both rotational and O 8. reflectional symmetry. 4. 5. P P Q O Q N All rights reserved. N 6. line symmetry and 72° rotational symmetry 9. Q' 7. P' S R' P line symmetry and 90° rotational symmetry Q 8. line symmetry and 45° rotational symmetry GuidedProblem Solving 9-3 1. The coordinates of point A, and three rotation transformations. It is assumed that the rotations are counterclockwise. 2. Parallelogram, rhombus, square 3. slope of A 5 2 y 5 © Pearson Education, Inc., publishing as Pearson Prentice Hall. R 9. line symmetry 10. A x line symmetryand 45° rotational symmetry O 11. 4. y 180° rotational symmetry B A x O 12. 13. COOK C D 60 Geometry All-In-One Answers Version B L1 Geometry: All-In-One Answers Version B (continued) 14. that the screen image has 16 16 = 256 as large an area as the figure on the transparency. 9. 2520 ft2 H OA X Practice 9-6 1. I. D II. C III. B IV. A 2. B B B Guided Problem Solving 9-4 m 1. the coordinates of one vertex of a figure that is symmetric about the y-axis 2. Line symmetry is the type of symmetry for which there is a reflection that maps a figure onto itself. 3. C C Practice 9-5 1. 5 3 2. 1 2 m C All rightsreserved. 3. the coordinates of another vertex of the figure 4. images (and preimages) 5. reflection across the y-axis 6. (-3, 4) 7. Yes 8. (-6, 7) 3. 2 4. yes 5. no 6. no m R 7. R 8. A J m J T J A 4. R m R y 5. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4 A A 1. A description of a squareprojected onto a screen by an overhead projector, including the square’s area and the scale factor in relation to the square on the transparency. 2. The scale factor of a dilation is the number that describes the size change from an original figure to its image. 3. the area of the square on the transparency 4.smaller; The scale factor 16 > 1, so the dilation is an enlargement. 5. 1 ; 1 6. 1 ; 16 16 256 1 1 Being as high and as wide, the square on the 16 16 tranparency has 1 3 1 5 1 times the area. 7. 3 ft2 16 16 256 256 8. The shape does not matter. Regardless of the shape, the figure is being enlarged by afactor of 16 in two directions, so All-In-One Answers Version B O 2 2 6 x 4 T T Guided Problem Solving 9-5 L1 2 T 9. P(-12, -12), Q(-6, 0), R(0, -6) 10. P(-2, 1), Q(-1, 0), R(0, 1) E S O B y 6. B 4 T 2 6 4 2 O 2 E 4 2 4 x S 7. reflection 8. rotation 9. glide reflection 10. translation Guided Problem Solving 9-61. assorted triangles and a set of coordinate axes 2. a transformation 3. the transformation that maps one Geometry 61 Geometry: All-In-One Answers Version B triangle onto another 4. They are congruent, which confirms that they could be related by one of the isometries listed. Corresponding vertices: Eand P, D and Q, C and M. 5. counterclockwise; clockwise. The triangles have opposite orientations, so the isometry must be a reflection or glide reflection. 6. No; there is no reflection line that will work. 7. a glide reflection consisting of a translation (x, y) → (x + 11, y) followed by a reflection across line y =0 (the x-axis) 8. Yes, both are horizontal. 9. 180º rotation with center Q 22 1 , 0 R 2 (continued) 4. 5. Possible answer: 6. Check students’ work. 7. Answers may vary. Sample: Practice 9-7 All rights reserved. 1. translational symmetry 2. 3. line symmetry, rotational symmetry, translational symmetry, glidereflectional symmetry 9A: Graphic Organizer 1. Transformations 2. Answers may vary. Sample: reflections; translations; compositions of reflections; and symmetry 3. Check students’ work. 9B: Reading Comprehension 1. D 2. E, R, 5, and 20 3. The center of the county is the center of the square milebounded by roads 12, 13, K and L. 4. C 5. 10 miles 6. a 4. 9C: Reading/Writing Math Symbols rotational symmetry, translational symmetry 5.–6. Samples: 5. 6. 1. reflection; opposite 2. rotation; same 3. glide reflection; opposite 4. translation; same 5. rotation; same 6. translation; same 9D: VisualVocabulary Practice/High-Use Academic Words 1. theorem 2. symbols 3. reduce 4. application 5. simplify 6. explain 7. characteristic 8. table 9. investigate 7. yes 8. no 9. yes Guided Problem Solving 9-7 1. two different-sized equilateral triangles and a trapezoid 2. A tessellation is a repeating pattern offigures that completely covers a plane, without gaps or overlaps. 3. Theorem 9-7 says that every quadrilateral tessellates. If the three polygons can be joined to form a quadrilateral (using each polygon at least once), that quadrilateral can be the basis of a tessellation. 62 Geometry 9E: Vocabulary CheckTransformation: A change in the position, size, or shape of a geometric figure. Preimage: The given figure before a transformation is applied. Image: The resulting figure after a transformation is applied. Isometry: A transformation in which an original figure and its image are congruent. Composition: Atransformation in which a second transformation is performed on the image of the first transformation. All-In-One Answers Version B L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. line symmetry, rotational symmetry, translational symmetry, glide reflectional symmetry Geometry: All-In-One Answers Version B (continued) 9F: Vocabulary Review Puzzle 1 2 P O G L Y H E T R Y T R Y D R O N I D 3 S Y M M E R 4 O B 5 T U S E F S 6 I S All rights reserved. E D M E C L T R L I O A O 13 T M L O E 10 15 E T A I T O I A N R A 14 L N N E 8 11 N S L O C U R C 9 A T L R T A T I N T S UF 16 O T I O R 12 D O I L A T I O N N D N E R 17 M 18 S C A V L E T N E R I T O E N X Chapter 10 © Pearson Education, Inc., publishing as Pearson Prentice Hall. C H I O 7 Practice 10-1 1. 8 2. 60 3. 1.92 4. 0.8125 5. 9 6. 1500 7. 8.75 8. 3.44 9. 30 10. 16 11. 12 12. 48 13. 40 Guided Problem Solving10-1 1. 21 bh 2. Check students’ work; B(7, 7) 3. a right angle Guided Problem Solving 10-3 1. 10 2. The angles are all congruent and the sides are all congruent. 3. 360 4. 36 5. bisects 6. 2 7. 18 8. 180 9. 72 10. 180 11. 30; 15; 75 Practice 10-4 1. 4 : 5; 16 : 25 2. 5 : 3; 25 : 9 3. 1 : 2 4. 8 : 3 5. 2 2 6. 2700 7cm 7. 1 : 2 8. 2 : 5 9. 108 ft 576 5 in.2 4. 7 5. OA 6. 7 7. 24.5 8. a square 9. 49 10. 12 11. 24.5 12. 21 b2 Guided Problem Solving 10-4 Practice 10-2 a < 20mm, b < 30mm and c < 40mm. 4. 40 mm 5 1 mm 5. < 90 mm 6. < 16 mm 7. < 320 mm2 8. 450 9. 8000 10. yes 11. 16,000 å 1. 48 cm2 2. 784 in.2 3.90 m2 4. 374 ft2 5. 160 cm2 6. 176.25 in.2 7. 42.5 square units 8. 226.2 in.2 1. scale 2. Check students’ work. 3. Answers will vary. Let a be the side opposite the 30° angle, b be the side opposite the 50° angle, and c be the side corresponding to 200 yd. Then 200 yd 5 yd Guided Problem Solving 10-2Practice 10-5 1. 12 h(b1 1 b2) 2. 3; 4; 2 3. 9 4. no 5. yes 6. 3 1. 174.8 cm2 2. 578 ft2 3. 1250.5 mm2 4. 1131.4 in.2 5. 37.6 m2 6. 30.1 mi2 7. 55.4 km2 8. 357.6 in.2 9. 9.7 mm2 10. 384.0 in.2 11. 83.1 m2 12. 65.0 ft2 13. 119.3 ft2 14. 8 m2 Practice 10-3 1. 3. 4. 6. x = 7; y = 3.5 " 3 2. x = 30; p = 4 " 3; q = 8m1 = 45; m2 = 45; m3 = 90; m4 = 90 m5 = 60; m6 = 30; m7 = 120 5. 120 in.2 137 in.2 L1 All-In-One Answers Version B Geometry 63 Geometry: All-In-One Answers Version B (continued) Guided Problem Solving 10-5 10D: Visual Vocabulary Practice 1. subtract 2. 50 3. 12 ap 4. 30 5. 4 6. tan4308 7. 1668. 116 9. yes 10. 151 1. area of a triangle 2. radius 3. altitude of a parallelogram 4. area of a trapezoid 5. area of a kite or rhombus 6. similarity ratio 7. apothem 8. perimeter (of an octagon) 9. height of a triangle 11 0 1. 16p 2. 7.8p 3. Samples: DFA, ABF 4. Samples: FE , 0 0 0 FD 5. Samples: FE , ED 6.32 7. 54 8. 71 9. 50 10. 140 11. 220 12. 130 13. 6p in. 14. 16p cm 10E: Vocabulary Check Guided Problem Solving 10-6 Apothem: The distance from the center to the side of a Semicircle: Half a circle. Adjacent arcs: Arcs on the same circle with exactly one point in common. 1. 11; 2 2. 11; 3; 33 3.circumference; 2 4. 31 5. 26 6. 41 7. 105 8. 102 9. 3027.15 regular polygon. Circumference: The distance around a circle. Concentric circles: Circles that lie on the same plane and 49 1. 49p 2. 21.2 3. 49 6 p 4. 6 p - 21.2 5. 4p 6. 7. 98 p 8. 23 p 9. 6p 10. 25 3 p 11. 3 12. 2 18 5p Guided Problem Solving 10-7 ? pr 2 2. 45 3. 25 4. 30 5. 27 6. a piece from 1. mAB 360 the outer ring 7. yes 8. a piece from the top tier 10F: Vocabulary Review Puzzle 1. adjacent arcs 2. semicircle 3. center 4. arc length 5. base 6. circle 7. central angle 8. congruent arcs 9. diameter 10. major arc 11. concentric circles 12. minor arc13. Pythagorean triple 14. radius 15. apothem 16. sector 17. height 18. circumference Highness, there is no royal road to geometry. Chapter 11 Practice 10-8 3 10 2 5 1. 8.7% 2. 10.9% 3. 15.3% 4. 5. 6. 1 7. 4.0% 8. 37.5% 9. 20% 10. 50% 11. 40% 12. 33 13 % Practice 11-1 Guided Problem Solving 10-8 1.8 2. 5 3. 12 4. C 5. D 6. B 7. A 8. triangle 9. rectangle 10. rectangle 1. outcomes 2. event 3. AF 4. FG 5. EG 6. AE; Check students’ work. 7. 2 or about 67% 8. 2 or about 3 3 67%; yes 9. 3 or 60% 5 10A: Graphic Organizer 1. Area 2. Answers may vary. Sample: areas of parallelograms and triangles; areasof trapezoids, rhombuses, and kites; areas of regular polygons; perimeters and areas of similar figures; trigonometry and area; circles and arcs; areas of circles and sectors; geometric probability 3. Check students’ work. 10B: Reading Comprehension 1. E 2. J 3. B 4. G 5. A 6. I 7. D 8. C 9. H 10. F 0 å å11. a2 + b2 = c2 12. mAC 5 1058 13. MN RS 0 0 14. "225 5 15 15. #JKL 16. XY > PQ 17. ~ABCD 18. 42 = 16 19. b 10C: Reading/Writing Math Symbols 1. 2860 ft2 2. 3331 ft2 3. 86 bundles 4. 2823 squares 5. $9.00 6. $27.00 7. 45 minutes 8. 21.5 hours 9. $645 10. $1419 64 Geometry All rights reserved.have the same center. Practice 10-7 Guided Problem Solving 11-1 1. a two-dimensional network 2. Verify Euler’s Formula for the network shown. 3. 4 4. 6; 9 5. 4 + 6 = 9 + 1; This is true. 6. F becomes 3, V stays at 6, E becomes 8. Euler’s Formula becomes 3 + 6 = 8 + 1, which is still true. 7. 4 + 4 = 6 + 2;This is true. Practice 11-2 1. 62.8 cm2 2. 226.2 cm2 3. 44.0 ft2 4. 84.8 cm2 5a. 320 m2 5b. 440 m2 6a. 576 in.2 6b. 684 in.2 7a. 48 cm2 7b. 60 cm2 8a. 1500 m2 8b. 1800 m2 9. 94.5p cm2 10. 290p ft2 Guided Problem Solving 11-2 1. a picture of a box, with dimensions 2. the amount of cardboard the boxis made of, in square inches 3. A front and a back, each 4 in. 7 1 in.; a top and a bottom, each 2 4 in. 1 in.; and one narrow side, 1 in. 7 1 in. 4. Front and 2 2 back: each 30 in. ; top and bottom: each 4 in.2; one narrow side: 7 1 in.2 5. 75 1 in.2 6. 75 1 in.2 is about half a square 2 2 2 foot (1 ft2 = 144 in.2),which is reasonable. The answer is an All-In-One Answers Version B L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Practice 10-6 Geometry: All-In-One Answers Version B (continued) © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. approximationbecause the solution did not consider the overlap of cardboard at the glued joints, nor does it factor in the trapezoid-shaped cutout in the front and back sides. 7. 3 21 in.2 Guided Problem Solving 11-6 1. the diameter of a hailstone, along with its density compared Practice 11-3 to normal ice 2. thehailstone’s weight 3. V 5 4 pr3 3 4. 2.8 in. 5. 91.95 in.3 6. 1.7 lb 7. The density of normal ice. This was given for comparison purposes only. 8. 0.01 oz 1. 620 cm2 2. 188 ft2 3. 36p cm2 4. 800p m2 5. 109.6 m2 6. 387.7 ft2 7. 118.4 cm2 8. 668.2 ft2 Practice 11-7 Guided Problem Solving 11-3 1. 7 : 9 2. 5 : 83. yes; 7 : 4 4. not similar 5. 125 cm3 6. 256 in.3 7. 25 ft2 8. 600 cm2 1. a solid consisting of a cylinder and a cone 2. the surface area of the solid 3. The top of the solid is a circular base of the cylinder, A1 = pr2. The side of the solid is the lateral area of the cylinder, A2 = 2prh. The bottom of the solid is thelateral area of the cone, A3 = pr. 4. r = 5 ft; h = 6 ft; Guided Problem Solving 11-7 or a little less than half. This seems reasonable. 8. 628 ft2 1. two different sizes for an image on a balloon, and the volume of air in the balloon for one of those sizes 2. the volume of air for the other size 3. The two sizes ofthe clown face have a ratio of 4 : 8, or 1 : 2. 4. 1 : 8 5. 108 in.3 8 = 864 in.3 6. Yes, the two clown face heights are lengths, measured in inches (not in.2 or in.3). 7. 351 cm3 Practice 11-4 11A: Graphic Organizer 1. 1131.0 m3 2. 7.1 in.3 3. 1781.3 in.3 4. 785.4 cm3 5. 120 in.3 6. 35 ft3 7. 1728 ft3 8. 265 in.39. 76 ft3 10. 1152 m3 1. Surface Area and Volume 2. Answers may vary. Sample: space figures and nets; space figures and drawing; surface areas; and volumes. 3. Check students’ work. Guided Problem Solving 11-4 11B: Reading Comprehension 1. dimensions for a layer of topsoil, and two options forbuying 1. Area = 12 base times height; area of a triangle 2. slope = the difference of the y-coordinates divided by the / 5 "52 1 122 5 13 ft 5. Top: A1 = 78.5 ft2; side: A2 = 188.5 ft2; bottom: 204.2 ft2 6. 471 ft2 7. About 0.43, the soil 2. which purchasing option is cheaper 3. 1 ft 3 4. 1400 ft3 5. 467 bags;$1167.50 6. 1400 ft3 < 52 yd3 7. $1164 (or slightly less if the topsoil volume is not rounded up to the next whole number) 8. buying in bulk 9. The two costs are quite close. It makes sense that buying in bulk would be slightly cheaper for a good-sized backyard. 10. Buying top soil by the bag would be lessexpensive because its would cost $335 and in bulk would cost at least $345.93. Practice 11-5 1. 34,992 cm3 2. 400 in.3 3. 10,240 in.3 4. 4800 yd3 5. 314.2 in.3 6. 4955.3 m3 7. 1415.8 in.3 8. 1005.3 m3 9. 20 10. 2 Guided Problem Solving 11-5 1. a description of a plumb bob, with dimensions 2. the plumbbob’s volume 3. equilateral; 2 cm 4. a = "3 cm; B = 6"3 cm2 5. Prism: V1 = 36"3 cm3; pyramid: V2 = 6"3 cm3; plum bob: V < 73 cm3 6. 1 ; this seems reasonable 7. 42 cm3 7 Practice 11-6 1. 2463.0 in.2 2. 6,157,521.6 m2 3. 12.6 cm2 4. 153.9 m2 5. 1436.8 mi3 6. 268,082.6 cm3 7. 7238.2 m3 8. 14.1 cm39. 5547 cm3 10. 42 cm3 L1 All-In-One Answers Version B difference of the x-coordinates; slope of the line between 2 points 3. a squared plus b squared equals c squared, the Pythagorean Theorem; the lengths of the sides of a right triangle 4. area = pi times radius squared; area of a circle 5. volume = pitimes radius squared times height; volume of a cylinder 6. the sum of the x-coordinates of two points divided by 2, and the sum of the y-coordinates of two points divided by 2; midpoint of a line segment with endpoints (x1, y1) and (x2, y2) 7. circumference = two times pi times the radius; circumference of acircle 8. volume = one-third pi times radius squared times height; volume of a cone 9. distance equals the square root of the quantity of the sum of the square of the difference of the x-coordinates, squared, plus the difference of the y-coordinates, squared; the distance between two points (x1, y1) and (x2,y2) 10. area = one-half the height times the sum of base one plus base two; area of a trapezoid 11. a 11C: Reading/Writing Math Symbols 1. 527.8 in2 2. 136 ft2 11D: Visual Vocabulary Practice 1. prism 2. cross section 3. sphere 4. cylinder 5. cone 6. pyramid 7. polyhedron 8. altitude 9. base Geometry 65Geometry: All-In-One Answers Version B (continued) 11E: Vocabulary Check Practice 12-3 Face: A surface of a polyhedron. Edge: An intersection of two faces of a polyhedron. Altitude: For a prism or cylinder, a perpendicular segment 1. A and D; B and C 2. ADB and CDB 3. 55 4. x = 45; y = 50; z = 855. x = 90; y = 70 6. 180 7. 70 8. x = 120; y = 100; z = 140 9a. mA = 90 9b. mB = 80 9c. mC = 90 9d. mD = 100 11F: Vocabulary Review 1. polyhedron 2. edge 3. vertex 4. cross section 5. prism 6. altitude 7. right prism 8. slant height 9. cone 10. volume 11. sphere 12. hemispheres 13. similar figures 14.hypotenuse Chapter 12 Guided Problem Solving 12-3 1. C 2. 180 3. a diameter 4. Check students’ work. 5. Check students’ work. 6. Check students’ work. 7. Check students’ work. 8. 90 9. Check students’ work. Practice 12-4 1. 87 2. 35 3. 120 4. 72 5. x = 58; y = 59; z = 63 6. x = 30; y = 30; z = 120 7. x =16; y = 52 8. x = 138; y = 111; z = 111 9. x = 30; y = 60 10. 10 11. 4 12. 3.2 13. 6 All rights reserved. that joins the planes of the bases. Surface area: For a prism, cylinder, pyramid, or cone, the sum of the lateral area and the areas of the bases. Volume: A measure of the space a figure occupies. Practice12-1 1. 32 2. 72 3. 15 4. 6 5. "634 6. 4 "3 7. circumscribed 8. inscribed 9. circumscribed Guided Problem Solving 12-1 1. The area of one is twice the area of the other. 2. inscribed; circumscribed 3. 4. Guided Problem Solving 12-4 1. tangent 2. 360. 3. 12 4. 85 5. 2 6. 180 7. 95 8. 104; 86; 75 9. 360 10. yes;The sum of the measures of the arcs should be 360. 11. Opposite angles are not supplementary. Practice 12-5 (0, 5) r 5. 2r 6. 4r2 7. "2r 8.2r 2 9. 4r 2 = 2(2r 2): One area is double the other area. 10.yes 11. The same: The area of (0, 0) one circle is twice the area of the other circle. Practice 12-2 0 GuidedProblem Solving 12-2 1. perpendicular 2. bisects 3. 4; 3 4. right 5. radius 6. Pythagorean Theorem 7. 5 8. yes 9. 8; 4 Geometry x 0 congruent chords, AB = BC = CA. Then, because an equilateral triangle is equiangular, mABC = mBCA = mCAB. 66 (5, 0) 5 1. r = 13; mAB 134.8 2. r = 3 "5; mAB 53.1 3. 3 4.4.5 5. 3 6.0 8.5 7. Q T; PR SU 0 8. A J; BC KL 9. Because congruent arcs have © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. C(0, 0), r = 6 2. C(-1, -6), r = 4 3. x2 + y2 = 49 4. (x - 5)2 + (y - 3)2 = 4 5. (x + 5)2 + (y - 4)2 = 14 6. (x + 2)2 + (y + 5)2 = 2 7. x2 + y2 = 4 8. x2 + (y - 3)2 = 16 9.(x - 7)2 + (y + 2)2 = 4 10. x2 + (y + 20)2 = 100 y 11. y 12. 5 4 3 2 1 3 (3, 5) 1 2 3 4 5 x All-In-One Answers Version B L1 Geometry: All-In-One Answers Version B (continued) Guided Problem Solving 12-5 Guided Problem Solving 12-6 1. (2, 2); 5 2. 43 3. They are perpendicular. 4. - 34 5. 39 4 6. y 5 234 x1 39 7. 13 8. y 5 3 x 1 39 ; yes 4 4 4 1. a line 2. Check students’ work. 3. They are perpendicular. 4. - 1 5. 2 6. the midpoint of PQ 7. (3, 2) 2 9. x-intercept: a = 25 ; y-intercept: b = 25 3 4 12A: Graphic Organizer Practice 12-6 1. Circles 2. Answers may vary. Sample: tangent lines; chords 1. and arcs;inscribed angles; and angle measures and segment lengths 3. Check students’ work. 12B: Reading Comprehension T Answers may vary. Sample answers: 1. AD, CE 2. BF, CD, AD 3. HD 4. AH 0 00 1 1 1 0 0 5. AF , AB , BC 6. CDF, DAB, DAC 7. AC , ED 1 8. /AOC 9. /DCE 10. DEA 11. /DAH 12. /AGB13. /AHC 14. /DCE 15. AH 16. a 1.5 cm All rights reserved. 8. -4 9. y = 2x - 4 10. (2, 0); yes 11. y 5 21 x 1 2 2. A 12C: Math Symbols * )Reading/Writing * ) B 1. BC ' DE 2. EF is tangent to (G. 3. WX > ZX 4. AB ' XY 5. HM = 4 3. 12D: Visual Vocabulary Practice Y X 4. B A 1. tangent to a circle 2. standardform of an equation of a circle 3. secant 4. chord 5. circumscribed about 6. intercepted arc 7. inscribed in 8. inscribed angle 9. locus © Pearson Education, Inc., publishing as Pearson Prentice Hall. 12E: Vocabulary Check Inscribed in: A circle inside a polygon with the sides of the E C polygon tangent tothe circle. Chord: A segment whose endpoints are on a circle. Point of tangency: The single point of intersection of a 5. a third line parallel to the two given lines and midway between them 6. a sphere whose center is the given point and whose radius is the given distance 7. the points within, but not on, acircle of radius 1 in. centered at the given point tangent line and a circle. Tangent to a circle: A line, segment, or ray in the plane of a circle that intersects the circle in exactly one point. Circumscribed about: circle outside a polygon with the vertices of the polygon on the circle. 8. 0.75 in. R L1 S 0.75 in. All-In-One Answers Version B 12F: Vocabulary Review 1. D 2. F 3. G 4. A 5. C 6. B 7. E 8. M 9. I 10. L 11. N 12. H 13. K 14. J Geometry 67 Geometry: All-In-One Answers Version B Name_____________________________________ Class____________________________ Date________________Lesson 1-1 Example. Patterns and Inductive Reasoning Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ 2 Using Inductive Reasoning Make a conjecture about the sum of the cubes NAEP 2005 Strand: GeometryUse inductive reasoning to make conjectures of the first 25 counting numbers. Find the first few sums. Notice that each sum is a perfect square and that the perfect squares form a pattern. Topic: Mathematical Reasoning Local Standards: ____________________________________ 3 1 3 1 3 1 3 1 3 1Vocabulary. conjecture is a conclusion you reach using inductive reasoning. A counterexample is an example for which the conjecture is incorrect. All rights reserved. A All rights reserved. Inductive reasoning is reasoning based on patterns you observe. half 2 2 96 2 48 the sum of the first 25 countingnumbers, or (1 ⫹ 2 ⫹ 3 ⫹ . . . ⫹ 25) 2. 2 the preceding term. The next two 24 and 24 2 12 . Quick Check. 1. Find the next two terms in each sequence. a. 1, 2, 4, 7, 11, 16, 22, 29 , 37 b. Monday, Tuesday, Wednesday, ,... , Thursday Friday ,... c. , ,... Answers may vary. Sample: Geometry Lesson 1-1 DailyNotetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. terms are 48 192 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Each term is All rights reserved. 2 1 2 (1 2) 2 (1 2 3) 2 (1 2 3 4) 2 (1 2 3 4 5) the sum of the cubes of the first 25 counting numbers equalsthe square of Use the pattern to find the next two terms in the sequence. 384, 192, 96, 48, . . . L1 Quick Check. 2. Make a conjecture about the sum of the first 35 odd numbers. Use your calculator to verify your conjecture. 1 1 1 3 4 22 1 3 5 9 32 1 3 5 7 16 42 1 3 5 7 9 25 52 12 The sum of the first 35 oddnumbers is 35 2, or 1225. L1 Daily Notetaking Guide Geometry Lesson 1-1 3 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date________________ Lesson 1-2 Example. Drawings, Nets, and Other Models 2 Drawing a Net Draw a net for the figure with a square base NAEP 2005 Strand: Geometry Topic: Dimension and Shape and four isosceles triangle faces. Label the net with its dimensions. 10 cm Lesson Objectives 1 Makeisometric and orthographic drawings 2 Draw nets for three-dimensional figures © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 1 2 9 3 2 36 6 2 100 10 2 225 15 The sum of the first two cubes equals the square of the sum of the first two counting numbers. The sum of the first threecubes equals the square of the sum of the first three counting numbers. This pattern continues for the fourth and fifth rows. So a conjecture might be that Example. 1 Finding and Using a Pattern Find a pattern for the sequence. 384 2 2 3 2 3 3 2 3 3 3 3 2 3 4 3 3 3 3 2 3 4 5 Think of the sides of the squarebase as hinges, and “unfold” the figure at these edges to form a net. The base of each of the four square isosceles triangle faces is a side of the . Write in the known dimensions. Local Standards: ____________________________________ Vocabulary. 8 cm orthographic drawing All rights reserved. onisometric dot paper. An is the top view, front view, and right-side view of a three- dimensional figure. A net is a two-dimensional pattern you can fold to form a three-dimensional figure. All rights reserved. An isometric drawing of a three-dimensional object shows a corner view of the figure drawn Example.8 cm 10 cm Quick Check. 1 Orthographic Drawing Make an orthographic drawing of the isometric drawing at right. Orthographic drawings flatten the depth of a figure. An orthographic three drawing shows views. Because no edge of the isometric drawing is hidden in the top, front, and right views, all linesare solid. 2. The drawing shows one possible net for the Graham Crackers box. ht t Rig on Front Top Right Quick Check. 1. Make an orthographic drawing from this isometric drawing. Front Top © Pearson Education, Inc., publishing as Pearson Prentice Hall. Fr © Pearson Education, Inc., publishing as

Pearson Prentice Hall. 14 cm 20 cm 7 cm AM AH GR KERS AC CR AM AH GR KERS AC CR 20 cm 7 cm 14 cm Draw a different net for this box. Show the dimensions in your diagram. Answers may vary. Example: 14 cm 20 cm Fr ht on t Rig Right 7 cm 4 L1 Geometry Lesson 1-2 Daily Notetaking GuideAll-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 1-2 Geometry 5 1 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 1-3 1 A plane is a flatsurface that has no thickness. Points, Lines, and Planes Lesson Objectives 2 Name_____________________________________ Class____________________________ Date ________________ NAEP 2005 Strand: Geometry Understand basic terms of geometry Understand basic postulates ofgeometry B C A Plane ABC Two points or lines are coplanar if they lie on the same plane. Topic: Dimension and Shape A postulate or axiom is an accepted statement of fact. Local Standards: ____________________________________ Examples. Vocabulary and Key Concepts. name three points thatare collinear and three points that are not collinear. Postulate 1-1 t Line t is the only line that passes through points A and B . B A Points All rights reserved. All rights reserved. Through any two points there is exactly one line. *AE ) B C E * , Z , and W lie on a line, so they are m Z Y 2 Using Postulate 1-4Shade the plane that contains X, Y, and Z. Postulate 1-2 A Y collinear. ) and BD intersect at C . Y V Postulate 1-4 Through any three noncollinear points there is exactly one plane. A point is a location. is the set of all points. Space A line is a series of points that extends in two opposite directions withoutend. t b. Name line m in three different ways. * )* )* ) Answers may vary. Sample: ZW , WY , YZ . 2. a. Shade plane VWX. Z b. Name a point that is coplanar with points V, W, and X. Y Y V X W are points that lie on the same line. Collinear points 6 B A no All rights reserved. ST Plane RST and plane STWintersect in ST . 1. Use the figure in Example 1. a. Are points W, Y, and X collinear? © Pearson Education, Inc., publishing as Pearson Prentice Hall. * ) R T W S © Pearson Education, Inc., publishing as Pearson Prentice Hall. If two planes intersect, then they intersect in exactly one line. X W QuickCheck. Postulate 1-3 W Z Points X, Y, and Z are the vertices of one of the four triangular faces of the pyramid. To shade the plane, shade the interior of the triangle formed by X , Y , and Z . If two lines intersect, then they intersect in exactly one point. D X 1 Identifying Collinear Points In the figure at right,Geometry Lesson 1-3 Daily Notetaking Guide L1 Daily Notetaking Guide L1 Geometry Lesson 1-3 7 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date ________________ Lesson 1-4 Examples. 1 Naming Segments and Rays Name the segments and rays in the figure. NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures The labeled points in the figure are A, B, and C. LocalStandards: ____________________________________ A segment is a part of a line consisting of two endpoints and all points between them. A segment is named by its two endpoints. So the BA (or AB) segments are and . BC (or CB) B A ray is a part of a line consisting of one endpoint and all thepoints of the line on one side of that endpoint. A ray is named by its endpoint first, followed ) . ) by any other point on the ray. So the rays are and BA BC Vocabulary. Segment AB AB A B A Endpoint Endpoint ) is the part of a line consisting of one endpoint and ray Ray YX YX all the points of the line onone side of the endpoint. X Y Endpoint All rights reserved. endpoints and all points between them. All rights reserved. A segment is the part of a line consisting of two ) RQ and ) R RS Parallel planes A J parallel ← → AB and CG are ← → to EF. skew lines. F are planes that do not intersect. G D AB is←→ G H B Plane ABCD is parallel to plane GHIJ. I C © Pearson Education, Inc., publishing as Pearson Prentice Hall. E ←→ . If the walls of your classroom ) ) 1. Critical Thinking Use the figure in Example 1. CB and BC form a line. Are they opposite rays? Explain. © Pearson Education, Inc., publishingas Pearson Prentice Hall. C B walls are parts of parallel planes. If the ceiling and opposite Quick Check. are opposite rays. Skew lines are noncoplanar; therefore, they are not parallel and do not intersect. H do not intersect are vertical, S are coplanar lines that do not intersect. D Planes are parallel if theyfloor of the classroom are level, they are parts of parallel planes. Q endpoint. A C 2 Identifying Parallel Planes Identify a pair of parallel planes in your classroom. Opposite rays are two collinear rays with the same Parallel lines A © Pearson Education, Inc., publishing as Pearson Prentice Hall. Segments,Rays, Parallel Lines and Planes Lesson Objectives 1 Identify segments and rays 2 Recognize parallel lines No; they do not have the same endpoint. 2. Use the diagram to the right. a. Name three pairs of parallel planes. PSWT RQVU, PRUT SQVW, PSQR TWVU S Q R P W * ) b. Name a line that isparallel to PQ . T V U * ) TV c. Name a line that is parallel to plane QRUV. * ) Answers may vary. Sample: PS 8 2 Geometry Lesson 1-4 Geometry Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 1-4 All-In-One Answers Version B 9 L1 Geometry: All-In-One Answers Version B(continued) Name_____________________________________ Class____________________________ Date________________ Lesson 1-5 Examples. Measuring Segments Lesson Objectives 1 Name_____________________________________ Class____________________________ Date________________ 1 Using the Segment Addition Postulate If AB 25, NAEP 2005 Strand: Measurement Find the lengths of segments 2x 6 find the value of x. Then find AN and NB. Local Standards: ____________________________________ NB AN ( Vocabulary and Key Concepts. )( 2x 6 AB 3x thatthe distance between any two points is the absolute value of the difference of the corresponding numbers. Simplify the left side. 24 Subtract 1 from each side. x 8 All rights reserved. All rights reserved. correspondence with the real numbers so Substitute. 1 25 3x ( )6 NB x 7 ( 8 ) 7 AN 2x 6 2 8 AN and NB10 15 B Segment Addition Postulate ) 25 x7 Postulate 1-5: Ruler Postulate The points of a line can be put into one-to-one x7 A N Use the Segment Addition Postulate (Postulate 1-6) to write an equation. Topic: Measuring Physical Attributes Divide each side by 3 . 10 Substitute 8 for x. 15 , which checksbecause the sum equals 25. Postulate 1-6: Segment Addition Postulate If three points A, B, and C are collinear and B B Find RM, MT, and RT. C 8x 36 5x 9 2 Finding Lengths M is the midpoint of RT. A is between A and C, then AB ⫹ BC ⫽ AC. R M T Use the definition of midpoint to write an equation. RMA coordinate is a point’s distance and direction from zero on a number line. All rights reserved. R AB 5 u a Q A B a b u b a coordinate of A coordinate of B Congruent (艑) segments are segments with the same length. 2 cm A B A D C 2 cm C B AB CD AB CD D midpoint © Pearson Education, Inc.,publishing as Pearson Prentice Hall. the length of AB © Pearson Education, Inc., publishing as Pearson Prentice Hall. 5x 9 5x segments. B | | AB MT ( RM 5x 9 5 ( MT 8x 36 8 15 15 Substitute. 8x RT RM MT 84 RM and MT are each 36 Add Subtract x )9 ) 36 midpoint Definition of 8x 36 45 3 x 15 to eachside. 5x from each side. Divide each side by 3 . 84 Substitute 84 168 15 for x. Segment Addition , which is half of 168 Postulate , the length of RT. Quick Check. 1. EG 100. Find the value of x. Then find EF and FG. 4x – 20 x ⫽ 15, EF ⫽ 40; FG ⫽ 60 A midpoint is a point that divides a segment into twocongruent A 45 E 2x + 30 F G C 2. Z is the midpoint of XY, and XY 27. Find XZ. BC 13.5 10 Geometry Lesson 1-5 Daily Notetaking Guide L1 Name_____________________________________ Class____________________________ Date________________ Lesson 1-6 © Pearson Education, Inc.,publishing as Pearson Prentice Hall. Examples. 1 Naming Angles Name the angle at right in four ways. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes m⬔ BOC A B m⬔AOC. 0 180 m⬔AOB m⬔BOC 180 T 1 Q ⬔TBQ 90 x5 x° x° x° angle 90 88 Substitute 42 for m⬔ABC. m⬔246 Subtract 42 Postulate for m⬔1 and 1 88 2 B from each side. C l2, lDEC A O C right Angle Addition m⬔2 1. a. Name ⬔CED two other ways. . B are the sides of the angle and the endpoint is the vertex of the angle. 0,x, 42 Quick Check. If ⬔AOC is a straight angle, then An angle (⬔) is formed by tworays with the same endpoint. The rays angle m⬔ 1 m⬔ 2 m⬔ABC B B acute A O O C x° All rights reserved. 14 0 30 15 0 20 16 0 17 0 10 D obtuse 90 angle ,x, straight angle x 5 180 180 © Pearson Education, Inc., publishing as Pearson Prentice Hall. AOB 13 0 50 y © Pearson Education, Inc.,publishing as Pearson Prentice Hall. m⬔ 12 0 60 10 Postulate 1-8: Angle Addition Postulate If point B is in the interior of ⬔AOC, then 70 17 0 A 80 x 20 . 90 1 A Use the Angle Addition Postulate (Postulate 1-8) to solve. 16 0 ) 7 0 10 0 6 0 11 0 20 50 0 1 3 10 0 1 10 All rights reserved. 40 C 80 0 15 0 3014 0 4 ) mlCOD 5 u x 2 y u 3 G 2 Using the Angle Addition Postulate Suppose that m⬔1 42 and m⬔ABC 88. Find m⬔2. b. If OC is paired with x and OD is paired with y, then . . Finally, the name can be a point on one side, the vertex, and a point on the other side of the angle: lAGC or lCGA . a. OA ispaired with 0 and OB is paired with 180 . ) lG The name can be the vertex of the angle: Vocabulary and Key Concepts. ) C l3 The name can be the number between the sides of the angle: Local Standards: ____________________________________ Postulate 1-7: Protractor Postulate ) ) Let OA and OBbe opposite rays in a plane. ) ) OA , OB , and all the rays with endpoint O that can * ) be drawn on one side of AB can be paired with the real numbers from 0 to 180 so that 11 Geometry Lesson 1-5 Name_____________________________________ Class____________________________ Date________________ Measuring Angles Lesson Objectives 1 Find the measures of angles 2 Identify special angle pairs Daily Notetaking Guide L1 b. Critical Thinking Would it be correct to name any of the angles ⬔E? Explain. No, 3 angles have E for a vertex, so you need more information in the name todistinguish them from one another. G 2. If m⬔DEG 145, find m⬔GEF. 35 D E F An acute angle has measurement between 0 and 90. A right angle has a measurement of exactly 90⬚. An obtuse angle has measurement between 90 and 180. A straight angle has a measurement of exactly 180⬚.Congruent angles are two angles with the same measure. 12 L1 Geometry Lesson 1-6 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 1-6 Geometry 13 3 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ 2 Constructing the Perpendicular Bisector Lesson 1-7 Basic Constructions Lesson Objectives Use a compass and a straightedge to construct congruent segments and congruent anglesUse a compass and a straightedge to bisect segments and angles Topic: Relationships Among Geometric Figures Local Standards: ____________________________________ same Step 2 With the compass setting, put the compass point on point B and draw another long arc. Label the intersect pointswhere the two arcs as X and Y. is a ruler with no markings on it. A compass is a geometric tool used to draw circles and parts of circles called arcs. Perpendicular lines are two lines that intersect to form right angles. A perpendicular bisector of a segment is a line, segment, or ray that is perpendicular tothe segment at its midpoint, thereby bisecting the segment A D C B All rights reserved. All rights reserved. Construction is using a straightedge and a compass to draw a geometric figure. straightedge J * ) T T W KM TW © Pearson Education, Inc., publishing as Pearson Prentice Hall. M Step 2 Open thecompass the length of KM. Step 3 With the same compass setting, put the compass point on point T. Draw an arc that intersects the ray. Label the point of intersection W. * ) Geometry Lesson 1-7 M B Y Quick Check. 1. Use a straightedge to draw XY. Then construct RS so that RS 2XY. X R Y DailyNotetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ S 2. Draw ST. Construct its perpendicular bisector. S 14 X A XY ⬜ AB at the midpoint of AB, so XY is the of AB. perpendicular bisector L K Step 1 Draw a raywith endpoint T. B * ) * ) Examples. to KM. X A K N 1 Constructing Congruent Segments Construct TW congruent B Step 3 Draw XY . The point of intersection of AB and XY is M, the of AB. midpoint © Pearson Education, Inc., publishing as Pearson Prentice Hall. coplanar angles. A Y into two congruentsegments. An angle bisector is a ray that divides an angle into two congruent B Step 1 Put the compass point on point A and draw a long arc. greater Be sure that the opening is than 12AB. Vocabulary. A A All rights reserved. 2 Given: AB. * ) * ) Construct: XY so that XY ⬜ AB at the midpoint M of AB.NAEP 2005 Strand: Geometry T Daily Notetaking Guide Geometry Lesson 1-7 15 Name_____________________________________ Class____________________________ Date ________________ 2 Finding an Endpoint The midpoint of DG is M(1, 5). One endpoint is Lesson 1-8 The CoordinatePlane Lesson Objectives 1 Find the distance between two points in the coordinate plane 2 Find the coordinates of the midpoint of a segment in the coordinate plane D(1, 4). Find the coordinates of the other endpoint G. Use the Midpoint Formula. Let (x 1, y 1 ) be NAEP 2005 Strand: Measurement Topic:Measuring Physical Attributes x 1x y 1y a 1 2 2 , 1 2 2 b be coordinates of G. Local Standards: ____________________________________ (1, 5) . Solve for (1, 4) x2 and the midpoint and y2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1Name_____________________________________ Class____________________________ Date ________________ , the Find the x-coordinate of G. d (x2 x1 )2 (y2 y1 )2 All rights reserved. Formula: The Distance Formula The distance d between two points A(x 1, y 1 ) and B(x 2, y 2 ) is . All rightsreserved. Key Concepts. ( x1 x2 ) y1 y2 , 2 2 1 x2 2 d Use the Midpoint Formula. S 5 2 1 x2 d Multiply each side by 2 . S 10 3 x2 Formula: The Midpoint Formula The coordinates of the midpoint M of AB with endpoints A(x 1, y 1 ) and B(x 2, y 2 ) are the following: M 1 d Simplify. S 6 4 y2 2 4 y2 x2 Thecoordinates of G are (3, 6) . Quick Check. . 1. Find the coordinates of the midpoint of XY with endpoints X(2, 5) and Y(6, 13). Examples. 1 Finding the Midpoint AB has endpoints (8, 9) and (6, 3). Find the coordinates of its midpoint M. Use the Midpoint Formula. Let (x 1, y 1 ) be (8, 9) and (x 2, y 2 ) be (6,3) . The midpoint has coordinates ( x1 x2 , y1 2 The x-coordinate is The y-coordinate is y2 ) 6 2 ) 1 Substitute 8 for x1 and 6 for x2. Simplify. 6 ) 3 Substitute 9 for y1 and 3 for y2. Simplify. (1, 3) . 2 8 ( 2 9 ( 2 3 2 The coordinates of the midpoint M are 16 4 Midpoint Formula . Geometry Lesson 1-8Geometry 2 Daily Notetaking Guide L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. (4, 4) 2. The midpoint of XY has coordinates (4, 6). X has coordinates (2, 3). Find the coordinates of Y. (6, 9) L1 Daily Notetaking GuideGeometry Lesson 1-8 All-In-One Answers Version B 17 L1 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________Class____________________________ Date ________________ Lesson 1-9 2 Finding Area of a Circle Find the area of 䉺B in terms of π. Perimeter, Circumference, and Area Lesson Objectives 1 Find perimeters of rectangles and squares, and circumferences of circles Find areas of rectangles,squares, and circles 2 NAEP 2005 Strand: Measurement In 䉺B, r Topic: Measuring Physical Attributes Aπ Local Standards: ____________________________________ Aπ ( A r dO h b Square with side length s. Perimeter P 4s s2 Rectangle with base b and height h. C Circle with radius r and diameterd. Perimeter P Circumference C Area A All rights reserved. b h s 2b ⫹ 2h All rights reserved. Perimeter and Area s yd. B Formula for the area of a circle 1.5 )2 2.25 π The area of 䉺B is Key Concepts. Area A 1.5 r2 Substitute 2.25p 1.5 1.5 yd for r. yd 2. Quick Check. 1. a. Find the circumference of a circlewith a radius of 18 m in terms of π. 36p m pd or C 2pr bh Area b. Find the circumference of a circle with a diameter of 18 m to the nearest tenth. pr 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. equal If two figures are congruent, then their areas are . All rights reserved. Postulate 1-10The area of a region is the sum of the areas of its non-overlapping parts. Examples. 1 Finding Circumference 䉺G has a radius of 6.5 cm. Find the circumference of 䉺G in terms of π. Then find the circumference to the nearest tenth. C2 p ( 6.5 C 13 π C 13 C 2π Formula for circumference of a circle )Substitute 6.5 2. You are designing a rectangular banner for the front of a museum. The banner will be 4 ft wide and 7 yd high. How much material do you need in square yards? 2 91 3 yd for r. Exact answer 40.840704 The circumference of 䉺G is 18 6.5 cm G r © Pearson Education, Inc., publishing asPearson Prentice Hall. 56.5m Postulate 1-9 13p Use a calculator. , or about 40.8 cm. Geometry Lesson 1-9 Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 1-9 19 Name_____________________________________ Class____________________________Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 2-1 2 Writing the Converse of a Conditional Write the converse of the following Conditional Statements conditional. © Pearson Education, Inc.,publishing as Pearson Prentice Hall. Lesson Objectives 1 Recognize conditional statements 2 Write converses of conditional statements NAEP 2005 Strand: Geometry Topic: Mathematical Reasoning If x 9, then x 3 12. The converse of a conditional exchanges the hypothesis and the conclusion. LocalStandards: ____________________________________ Conditional Vocabulary and Key Concepts. If an angle is a straight angle, Converse If the measure of an angle is Symbolic Form If p , then q . then its measure is 180. If q , qSp 180, then it is a straight angle. Conclusion Hypothesis Conclusion x9 x3 12 x ⫹ 3 ⫽ 12 x⫽9 So the converse is: If x ⫹ 3 ⫽ 12, then x ⫽ 9. You read it pSq All rights reserved. Conditional All rights reserved. Conditional Statements and Converses Statement Example Converse Hypothesis then p . Quick Check. 1. Identify the hypothesis and the conclusion of this conditionalstatement: If y 3 5, then y 8. Hypothesis: A conditional is an if-then statement. y⫺3⫽5 hypothesis is the part that follows if in an if-then statement. truth value The © Pearson Education, Inc., publishing as Pearson Prentice Hall. The conclusion is the part of an if-then statement (conditional) that follows then.of a statement is “true” or “false” according to whether the statement is true or false, respectively. The converse of the conditional “if p, then q” is the conditional “if q, then p.” Examples. 1 Identifying the Hypothesis and the Conclusion Identify the hypothesis and conclusion: If two lines are parallel, then thelines are coplanar. In a conditional statement, the clause after if is the hypothesis and the clause after then is the conclusion. © Pearson Education, Inc., publishing as Pearson Prentice Hall. The Conclusion: y⫽8 2. Write the converse of the following conditional: If two lines are not parallel and do notintersect, then they are skew. If two lines are skew, then they are not parallel and do not intersect. Hypothesis: Two lines are parallel. Conclusion: The lines are coplanar. 20 L1 Geometry Lesson 2-1 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 2-1Geometry 21 5 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ 2 Identifying a Good Definition Is the following statement a good Lesson 2-2 Biconditionals and Definitions LessonObjectives 1 definition? Explain. An apple is a fruit that contains seeds. NAEP 2005 Strand: Geometry Write biconditionals Recognize good definitions 2 Name_____________________________________ Class____________________________ Date ________________ The statement is true as adescription of an apple. Topics: Dimension and Shape; Mathematical Reasoning Exchange “An apple” and “a fruit that contains seeds.” The converse reads: Local Standards: ____________________________________ A fruit that contains seeds is an apple. Vocabulary and Key Concepts. There aremany fruits that contain seeds but are not apples, such as lemons and peaches. These are Biconditional Statements A biconditional combines p S q and q S p as p 4 q. Symbolic Form An angle is a straight angle if You read it p if and p4q only if q . and only if its measure is 180°. All rights reserved.Example Biconditional All rights reserved. Statement statement is biconditional The original statement statement is not counterexamples , so the converse of the . is not a good definition because the reversible. Quick Check. 1. Consider the true conditional statement. Write its converse. If the converse isalso true, combine the statements as a biconditional. A biconditional statement is the combination of a conditional statement and its converse. A false contains the words “if and only if.” Conditional: If three points are collinear, then they lie on the same line. Converse: Examples. If three points lie on thesame line, then they are collinear. 1 Writing a Biconditional Consider the true conditional statement. Write its To write the converse, exchange the hypothesis and conclusion. Converse: If x ⫹ 15 ⫽ 20, then x ⫽ 5. When you subtract 15 from each side to solve the equation, you get x 5. Because true boththe conditional and its converse are a biconditional using the phrase , you can combine them in if and only if . Biconditional: x ⫽ 5 if and only if x ⫹ 15 ⫽ 20. 22 Geometry Lesson 2-2 Daily Notetaking Guide L1 true The converse is . Biconditional: All rights reserved. Conditional: If x 5, then x 15 20. ©Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. converse. If the converse is also true, combine the statements as a biconditional. Three points are collinear if and only if they lie on the same line. 2. Is the following statement agood definition? Explain. A square is a figure with four right angles. It is not a good definition because a rectangle has four right angles and is not necessarily a square. L1 Daily Notetaking Guide Geometry Lesson 2-2 23 Name_____________________________________Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 2-3 3 Using the Law of Syllogism Use the Law of Syllogism to draw a Deductive Reasoning conclusion fromthe following true statements: If a quadrilateral is a square, then it contains four right angles. If a quadrilateral contains four right angles, then it is a rectangle. NAEP 2005 Strand: Geometry Topic: Mathematical Reasoning Local Standards: ____________________________________ The conclusion ofthe first conditional is the hypothesis of the second conditional. This means that you can apply the Vocabulary and Key Concepts. The Law of Syllogism: If then is true. All rights reserved. conclusion In symbolic form: If p S q is a true statement and p is true, then q is true. Law of Syllogism All rightsreserved. Law of Detachment If a conditional is true and its hypothesis is true, then its © Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson Objectives 1 Use the Law of Detachment 2 Use the Law of Syllogism pSr pSq and qSr Law of Syllogism . are true statements, is a true statement.So you can conclude: If a quadrilateral is a square, then it is a rectangle. Quick Check. 1. Suppose that a mechanic begins work on a car and finds that the car will not start. Can the mechanic conclude that the car has a dead battery? Explain. If p S q and q S r are true statements, then p S r is a truestatement. No, there could be other things wrong with the car, such as a faulty starter. Deductive reasoning is a process of reasoning logically from given facts to a conclusion. 1 Using the Law of Detachment A gardener knows that if it rains, the garden will be watered. It is raining. What conclusion can hemake? The first sentence contains a conditional statement. The hypothesis is it rains. Because the hypothesis is true, the gardener can conclude that the garden will be watered. 2 Using the Law of Detachment For the given statements, what can you conclude? Given: If ⬔A is acute, then m⬔A 90°. ⬔A isacute. A conditional and its hypothesis are both given as true. By the Law of Detachment conclusion of the conditional, m⬔A 90°, is 24 6 Geometry Lesson 2-3 Geometry © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall.Examples. 2. If a baseball player is a pitcher, then that player should not pitch a complete game two days in a row. Vladimir Nuñez is a pitcher. On Monday, he pitches a complete game. What can you conclude? Answers may vary. Sample: Vladimir Nuñez should not pitch a complete game on Tuesday.3. If possible, state a conclusion using the Law of Syllogism. If it is not possible to use this law, explain why. a. If a number ends in 0, then it is divisible by 10. If a number is divisible by 10, then it is divisible by 5. If a number ends in 0, then it is divisible by 5. b. If a number ends in 6, then it is divisible by 2.If a number ends in 4, then it is divisible by 2. , you can conclude that the true Not possible; the conclusion of one statement is not the hypothesis of the other statement. . Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 2-3 All-In-One Answers Version B 25 L1 Geometry: All-In-OneAnswers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 2-4 Examples. Reasoning in Algebra Lesson Objective 1 Name_____________________________________Class____________________________ Date ________________ 1 Justifying Steps in Solving an Equation Justify each step used to solve NAEP 2005 Strand: Algebra and Geometry Connect reasoning in algebra and geometry 5x 12 32 x for x. Topics: Algebraic Representations; MathematicalReasoning Given: 5x 12 32 x Local Standards: ____________________________________ 5x 44 x Addition Property of Equality 4x 44 Subtraction Property of Equality Key Concepts. x 11 If a b, then a c b c . Multiplication Property If a b, then a c b c . If a b and c 0, then b c If a b, then b a . TransitiveProperty If a b and b c, then a c . Substitution Property If a b, then b can replace a in any expression. Distributive Property a(b c) ac . AB ⬔A ⬔A If AB CD, then CD AB . If ⬔A ⬔B, then ⬔B ⬔A . If AB CD and CD EF , then AB Transitive Property ⬔P 艑 ⬔S If ⬔P ⬔R and ⬔R ⬔S, then Properties ofCongruence Symmetric Property . for ⬔P ⬔R and the third part of the hypothesis: . EF © Pearson Education, Inc., publishing as Pearson Prentice Hall. ab ⬔P 艑 ⬔R Transitive Property of Congruence Use the a a . for the first two parts of the hypothesis: If ⬔P ⬔Q and ⬔Q ⬔R, then Symmetric PropertyAB Transitive Property of Congruence Use the . Reflexive Property Reflexive Property If ⬔P ⬔Q, ⬔Q ⬔R, and ⬔R ⬔S, then ⬔P ⬔S. c © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. a justifies each statement. All rights reserved. If a b, then a c b c . SubtractionProperty All rights reserved. Addition Property Division Property Division Property of Equality 2 Using Properties of Equality and Congruence Name the property that Properties of Equality . Quick Check. M 1. Fill in each missing reason. ) (2x40) Given: LM bisects ⬔KLN. ) LM bisects ⬔KLN 4x Given Km⬔MLN m⬔KLM Definition of Angle 4x 2x 40 Substitution Property of Equality 2x 40 Subtraction Property of Equality x 20 N L Bisector Division Property of Equality 2. Name the property of equality or congruence illustrated. a. XY XY Reflexive Property of Congruence If ⬔A ⬔B and ⬔B ⬔C, then ⬔A⬔C . b. If m⬔A 45 and 45 m⬔B, then m⬔A m⬔B Transitive or Substitution Property of Equality 26 Geometry Lesson 2-4 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Lesson 2-5 27 GeometryLesson 2-4 Name_____________________________________ Class____________________________ Date ________________ complementary angles Proving Angles Congruent Lesson Objectives 1 Prove and apply theorems about angles © Pearson Education, Inc., publishing as Pearson PrenticeHall. Daily Notetaking Guide NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures supplementary angles 2 4 3 1 Local Standards: ____________________________________ ⬔1 and ⬔2 are complementary angles. Vocabulary and Key Concepts. Theorem 2-1: Vertical AnglesTheorem 3 4 2 ⬔1 ⬔2 and ⬔3 ⬔4 Theorem 2-2: Congruent Supplements Theorem If two angles are supplements of the same angle (or of congruent angles), then the two angles are congruent All rights reserved. 1 . All rights reserved. congruent Vertical angles are . congruent Two angles arecomplementary angles if Two angles are supplementary angles if the sum of their measures is 90°. the sum of their measures is 180°. A theorem is a conjecture that is proven. A paragraph proof is a convincing argument that uses deductive reasoning in which statements and reasons are connected insentences. Examples. 1 Using the Vertical Angles Theorem Find the value of x. The angles with labeled measures are vertical angles. Apply the Vertical Angles Theorem to find x. Theorem 2-3: Congruent Complements Theorem If two angles are complements of the same angle (or of congruent angles),then the two angles are 4x 101 2x 3 . angle. right adjacent angles 1 3 4 2 ⬔1 and ⬔2 are vertical angles, as are ⬔3 and 3 1 ⬔4 . 2 4 ⬔1 and ⬔2 are adjacent angles, as are ⬔3 and ⬔4 . Vertical angles are two angles whose Adjacent angles are two coplanar angles sides form two pairs of opposite rays.that have a common side and a common © Pearson Education, Inc., publishing as Pearson Prentice Hall. vertical angles © Pearson Education, Inc., publishing as Pearson Prentice Hall. angles are congruent. Theorem 2-5 If two angles are congruent and supplementary, then each is a (2x 3) (4x 101)Vertical Angles Theorem 4x 2x 104 Theorem 2-4 right All ⬔3 and ⬔4 are supplementary angles. Addition Property of Equality 2x 104 Subtraction Property of Equality x 52 Division Property of Equality 2 Proving Theorem 2-2 Write a paragraph proof of Theorem 2-2 1 using the diagram at the right. Startwith the given: ⬔1 and ⬔2 are supplementary, ⬔3 and ⬔2 are supplementary. By the definition of supplementary angles 2 3 , m⬔1 m⬔2 180 and m⬔3 m⬔2 180. By substitution, m⬔1 m⬔2 subtract m⬔2 m⬔2 m⬔3 . Using the from each side. You get m⬔1 Subtraction Property of Equality m⬔3 , , or⬔ 1 ⬔ 3 . Quick Check. 1. Refer to the diagram for Example 1. a. Find the measures of the labeled pair of vertical angles. 107° b. Find the measures of the other pair of vertical angles. vertex but no common interior points. 73° c. Check to see that adjacent angles are supplementary. 107° ⫹ 73° ⫽ 180° 28L1 Geometry Lesson 2-5 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 2-5 Geometry 29 7 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________Date________________ Lesson 3-1 Properties of Parallel Lines Lesson Objectives 1 Examples. NAEP 2005 Strand: Measurement Identify angles formed by two lines and a transversal Prove and use properties of parallel lines 2 Name_____________________________________Class____________________________ Date ________________ 1 Applying Properties of Parallel Lines In the diagram of Topic: Measuring Physical Attributes Lafayette Regional Airport, the black segments are runways and the gray areas are taxiways and terminal buildings. Local Standards:____________________________________ Compare ⬔2 and the angle vertical to ⬔1. Classify the angles as alternate interior angles, same-side interior angles, or corresponding angles. Vocabulary and Key Concepts. 2 3 1 / m ⬔1 ⬔2 a t b 3 2 1 are nonadjacent interior angles that t 56 13 42 78 lie onopposite sides of the transversal. Same-side interior angles are interior angles that lie on the same side of the transversal. ⬔1 and ⬔2 are alternate interior angles. of the transversal and in corresponding positions relative to the corresponding angles Because ᐉm, m⬔1 by the 42 Angle Addition Postulatefor m⬔1, the equation becomes Subtract p q . Daily Notetaking Guide 4 . 5 42 / m 42 42 . If you substitute 42 m⬔2 180 . from each side to find m⬔2 138 . 1. Use the diagram in Example 1. Classify ⬔2 and ⬔3 as alternate interior angles, same-side interior angles, or corresponding angles. same-sideinterior angles 2. Using the diagram in Example 2 find the measure of each angle. Justify each answer. a. ⬔3 42; Vertical angles are congruent b. ⬔4 138; Corresponding angles are congruent c. ⬔5 138; Same-side interior angles are supplementary L1 L1Name_____________________________________ Class____________________________ Date________________ Lesson 3-2 8 6 7 2 1 3 Quick Check. ⬔1 and ⬔7 corresponding angles. coplanar lines. Daily Notetaking Guide Geometry Lesson 3-1 31Name_____________________________________ Class____________________________ Date ________________ Examples. Proving Lines Parallel 1 Proving Theorem 3-5 If two lines and a transversal form alternate NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes 3 / interiorangles that are congruent, then the two lines are parallel. Given: ⬔1 ⬔2 Prove: ᐉ m Local Standards: ____________________________________ 1 m 2 Write the flow proof below of the Alternate Interior Angles Theorem as a paragraph proof. Vocabulary and Key Concepts. ᐉ m . Theorem 3-5:Converse of the Alternate Interior Angles Theorem If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel. ⬔1 ⬔3 All rights reserved. then the two lines are parallel. 2 All rights reserved. m If two lines and a transversal form corresponding angles thatare congruent, ⬔1 ⬔2 Given 1 / Postulate 3-2: Converse of the Corresponding Angles Postulate Vertical angles are congruent ⬔3 then ᐉ m . congruent, then the two lines are parallel. Theorem 3-8: Converse of the Same-Side Exterior Angles Theorem If ⬔3 and ⬔6 are If two lines and a transversal formsame-side exterior angles that are supplementary, supplementary, then the two lines are parallel. then ᐉ m . A flow proof uses arrows to show the logical connections between the statements. © Pearson Education, Inc., publishing as Pearson Prentice Hall. If two lines and a transversal form alternateexterior angles that are by the ⬔3 and Transitive ⬔2 Property of Congruence. are corresponding angles, ᐉ m by the Postulate. Use the diagram at the right. Which lines, if any, must be parallel if ⬔3 and ⬔2 are supplementary? Justify your answer with a theorem or postulate. © Pearson Education, Inc.,publishing as Pearson Prentice Hall. supplementary, If ⬔3 ⬔5, ? 2 Using Postulate 3-2 If two lines and a transversal form same-side interior angles that are then ᐉ m . / || m Converse of the Corresponding Angles then ᐉ m . supplementary, then the two lines are parallel. ⬔2 Because If ⬔2 and ⬔4 areTheorem 3-7: Converse of the Alternate Exterior Angles Theorem ⬔3 ⬔2 Transitive Property of Congruence By the Vertical Angles Theorem, ⬔3 ⬔1 . ⬔1 ⬔2, so 5 6 1 4 2 m 3 If ⬔1 ⬔2, / Theorem 3-6: Converse of the Same-Side Interior Angles Theorem Reasons . straight angle, m⬔1 m⬔2 180 by the⬔1 and ⬔4 are same-side interior angles. are angles that lie on the same side Geometry Lesson 3-1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. / © Pearson Education, Inc., publishing as Pearson Prentice Hall. 180 m Lesson Objectives 1 Use a transversal in proving lines parallel⬔1 and the 42 angle are Because ⬔1 and ⬔2 are adjacent angles that form a distinct points. 30 alternate interior angles Corresponding Angles Postulate A transversal is a line that intersects two coplanar lines at two Corresponding angles ⬔2 and the angle vertical to ⬔1 are In the diagram at right, ᐉmand pq. Find m⬔1 and m⬔2. Theorem 3-2: Same-Side Interior Angles Theorem If a transversal intersects two parallel lines, then same-side interior angles are supplementary . m⬔1 m⬔2 are not adjacent and lie between the lines on opposite sides of the transversal, ⬔3 ⬔1 alternate interior anglestransversal runway. Because 2 Finding Measures of Angles Theorem 3-1: Alternate Interior Angles Theorem If a transversal intersects two parallel lines, then alternate interior congruent angles are . Alternate interior angles ⬔2 is between the runway segments and on the opposite side of the All rightsreserved. 1 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. t All rights reserved. Postulate 3-1: Corresponding Angles Postulate If a transversal intersects two parallel lines, then corresponding congruent angles are . All rights reserved. The angle vertical to ⬔1 is between the runwaysegments. E It is given that ⬔3 and ⬔2 are supplementary. The diagram shows that ⬔4 and ⬔2 are supplementary. Because supplements of the same angle are 3 C 4K 2 congruent (Congruent Supplements Theorem), ⬔3 and D1 ⬔4 . Because ⬔3 ⬔4 are congruent corresponding angles, ECConverse of the Corresponding Angles ) DK ) by the Postulate. Quick Check. 1. Supply the missing reason in the flow proof from Example 1. If corresponding angles are congruent, then the lines are parallel. 2. Use the diagram from Example 1. Which lines, if any, must be parallel if ⬔3 ⬔4? Explain. arewritten below the statements. u u EC 储 DK ; Converse of Corresponding Angles Postulate 32 8 Geometry Lesson 3-2 Geometry Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 3-2 All-In-One Answers Version B 33 L1 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ Lesson 3-3 Examples. Parallel and Perpendicular Lines Lesson Objectives 1 Name_____________________________________ Class____________________________ Date________________ 1 Proof of Theorem 3-10 NAEP 2005 Strand: Geometry Relate parallel and perpendicular lines Local Standards: ____________________________________ Key Concepts. By the Vertical Angles Theorem, ⬔1 is Because ⬔1 they are parallel a b , then All rights reserved. parallel tothe same line If two lines are c to each other. a b . perpendicular to the same line In a plane, if two lines are then they are t m Theorem 3-10 parallel , ⬔1 by the Converse of the Alternate Interior Angles and the Theorem to prove that the lines are parallel. 2 Using Algebra n m n . Find the value of x forwhich ᐉ m. The labeled angles are alternate interior angles . If ᐉm, the alternate interior angles are , equal and their measures are m All rights reserved. m 5x 66 5x equation ᐉ © Pearson Education, Inc., publishing as Pearson Prentice Hall. n © Pearson Education, Inc., publishing as Pearson PrenticeHall. Theorem 3-11 In a plane, if a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other. ⊥ congruent (14 3x) . Write and solve the 14 3x 5x 80 3x 2x 80 Daily Notetaking Guide L1 Subtract 3x from each side. x 40 Divide each side by 2. Quick Check. 1. Critical ThinkingIn a plane, if two lines form congruent angles with a third line, must the lines be parallel? Draw a diagram to support your answer. No; answers may vary. Sample: 115° 115° L1 Name_____________________________________ Class____________________________ Date________________ ©Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson Objectives 1 Classify triangles and find the measures of their angles 2 Use exterior angles of triangles Daily Notetaking Guide Examples. 48 m⬔Z 67 180 m⬔Z 180 115 Local Standards: ____________________________________ 21 measures of its two remote interior angles. Back B A All rights reserved. . All rights reserved. C Theorem 3-13: Triangle Exterior Angle Theorem The measure of each exterior angle of a triangle equals the sum of the 65 x 67° Y X 48° Subtract 115 from each side. 3 x (180180 Arm ) x adjacent anglesstraight angle measure decreases three acute angles. one right angle. __________________ __________________ __________________ __________________ An equilateral triangle has An equiangular triangle has one obtuse angle. __________________ __________________ three congruentangles. ______________________ ______________________ An isosceles triangle has A scalene triangle has at least two congruent sides. __________________________ __________________________ no congruent sides. _____________________ _____________________ An exterior angle of apolygon is an angle formed by a side and an extension of an adjacent side. Remote interior All-In-One Answers Version B increases as you make a lounge chair recline more. 32 b. Reasoning Explain why this statement must be true: If a triangle is a right triangle, its acute angles are complementary. Thesum of the measures of the angles of a triangle is 180. If you subtract the measure of the right angle from 180, you get 90. The sum of the measures of the other two angles is 90, so they are complementary. 2. a. Find m⬔3. 90 45 b. Critical Thinking Is it true that if two acute angles of a triangle arecomplementary, then the triangle must be a right triangle? Explain. 45 3 True, because the measures of the two complementary angles add to 90, leaving 90 for the third angle. angles Daily Notetaking Guide , which together form a . As one measure increases, the other . The angle formed by the back ofthe chair and 1. a. 䉭MNP is a right triangle. ⬔M is a right angle and m⬔N is 58. Find m⬔P. 2 3 ) x Quick Check. angle Remote interior angles are the two nonadjacent interior angles corresponding to each exterior angle of a triangle. Exterior 1 © Pearson Education, Inc., publishing as Pearson PrenticeHall. An obtuse triangle has © Pearson Education, Inc., publishing as Pearson Prentice Hall. the armrest A right triangle has (180180 The exterior angle and the angle formed by the back of the chair and the armrest are L1 Simplify. the angle formed by the back of the chair and the armrest as you make alounge chair recline more. 180 m⬔A m⬔B m⬔C 180 Geometry Lesson 3-4 Z Triangle Angle-Sum Theorem 2 Applying the Triangle Exterior Angle Theorem Explain what happens to Theorem 3-12: Triangle Angle-Sum Theorem The sum of the measures of the angles of a triangle is 36 35 GeometryLesson 3-3 m⬔Z three congruent sides. ________________________ ________________________ (7x 8) 62 1 Applying the Triangle Angle-Sum Theorem Find m⬔Z. NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures Vocabulary and Key Concepts. An acute triangle has a bName_____________________________________ Class____________________________ Date ________________ Parallel Lines and the Triangle Angle-Sum Theorem Lesson 3-4 m Add 66 to each side. x 18; 7(18) 8 118°, and 62° 118° 180° Geometry Lesson 3-3 / (5x 66) 3x. 14 66 2. Find the valueof x for which a b. Explain how you can check your answer. 34 t 2 1 Property of Congruence. Because alternate interior angles are congruent, you can use the vertical angle of ⬔1 to each other. n r to its vertical angle. congruent ⬔2 , ⬔2 is congruent to the vertical angle of Transitive All rights reserved.Theorem 3-9 s Use the diagram at the right. Which angle would you use with ⬔1 to prove the theorem In a plane, if two lines are perpendicular to the same line, then they are parallel to each other (Theorem 3-10) using the Converse of the Alternate Interior Angles Theorem instead of the Converse of theCorresponding Angles Postulate? Topic: Relationships Among Geometric Figures L1 L1 Daily Notetaking Guide Geometry Lesson 3-4 Geometry 37 9 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________Date________________ Lesson 3-5 Name_____________________________________ Class____________________________ Date ________________ An equilateral polygon has all sides congruent. The Polygon Angle-Sum Theorems An equiangular polygon has all angles congruent . LessonObjectives 1 NAEP 2005 Strand: Geometry Classify polygons Find the sums of the measures of the interior and exterior angles of polygons 2 A Topic: Relationships Among Geometric Figures is both equilateral and equiangular. regular polygon Local Standards:____________________________________ Examples. 1 Classifying Polygons Classify the polygon at the right by its sides. Identify it as convex or concave. Vocabulary and Key Concepts. Starting with any side, count the number of sides clockwise around the figure. Because the polygon has 12 sides, itis a dodecagon. Think of the . Theorem 3-15: Polygon Exterior Angle-Sum Theorem The sum of the measures of the exterior angles of a polygon, polygon as a star. Draw a diagonal connecting two adjacent points of the star. That diagonal lies 2 4 360 2 Finding a Polygon Angle Sum Find the sum of themeasures of the angles 1 . 5 of a decagon. A polygon is a closed plane figure with at least three sides that are segments. The sides B A C E D C D D Not a polygon; A polygon not a closed E Not a polygon; figure two sides intersect between endpoints. A convex polygon does not have diagonal points Aoutside of the polygon. D R Y A convex polygon A concave polygon has at least one diagonal T Diagonal P M K W Q G with points outside of the polygon. Geometry Lesson 3-5 ( ) 2 (180) 10 . Polygon Angle-Sum Substitute 8 ? 180 Subtract. Simplify. 1440 10 Theorem for n. Quick Check. 1. Classify each

polygon by its sides. Identify each as convex or concave. a. b. hexagon; convex octagon; concave 2. Find the sum of the measures of the angles of a 13-gon. 1980 A concave polygon S 38 © Pearson Education, Inc., publishing as Pearson Prentice Hall. E © Pearson Education, Inc., publishing asPearson Prentice Hall. B A C 10 Sum (n 2)(180) intersect only at their endpoints, and no two adjacent sides are collinear. B sides, so n 10 A decagon has All rights reserved. . For the pentagon, m⬔1 m⬔2 m⬔3 m⬔4 m⬔5 A the polygon, so the dodecagon is outside . concave Daily Notetaking Guide L1L1 Daily Notetaking Guide Geometry Lesson 3-5 39 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 3-62 Using Point-Slope Form Write an equation in point-slope form Lines in the Coordinate Plane Lesson Objectives 1 Graph lines given their equations 2 Write equations of lines of the line with slope 8 that contains P(3, 6). NAEP 2005 Strand: Algebra Topics: Patterns, Relations, and Functions; AlgebraicRepresentations y y Slope-intercept 1 Ax By C. Standard The point-slope form for a nonvertical line is form y ⫺ y 1 ⫽ m(x ⫺ x 1 ). 8 x1 (x (x 3 3 ) ) ) Use point-slope form. Substitute 8 for m and (3, 6) for (x1, y1). Simplify. 1. Graph each equation. form of a linear equation is standard O 4 2x y 1 4 3 x Allrights reserved. The y All rights reserved. y 2x 3 y ⫽ mx ⫹ b. ( m x ) Quick Check. form The slope-intercept form of a linear equation is y1 6 y 6 8 Local Standards: ____________________________________ Vocabulary. ( © Pearson Education, Inc., publishing as Pearson Prentice Hall. 360 one at eachvertex, is 3 All rights reserved. (n ⫺ 2)180 The sum of the measures of the angles of an n-gon is All rights reserved. Theorem 3-14: Polygon Angle-Sum Theorem a. y 12x 2 1 x O y 1 2 4 c. 5x y 3 2 x O 3 y 2 2(x 1) Point-slope b. 2x 4y 8 y O y 2 x 2 form Examples. 2. Write an equation of the line thatcontains the points P(5, 0) and Q(7, 3). 1 Graphing Lines Using Intercepts Use the x-intercept and y-intercept to 5x 6y 30 ( ) 6y 30 5 0 5x 0 30 0 6y 30 6y 30 x 6 30 y 5 The x-intercept is 6 . A point on the line is To find the y-intercept, substitute 0 for x and solve for y. 5x 6y 30 ( ) 30 5x 6 0 5x © PearsonEducation, Inc., publishing as Pearson Prentice Hall. To find the x-intercept, substitute 0 for y and solve for x. ( 6 , 0). The y-intercept is 5 . ( A point on the line is 0, 5 Plot (6, 0) and (0, 5). Draw the line containing the two points. ). y 2 (6, 0) 2 O 2 4 6 © Pearson Education, Inc., publishing as PearsonPrentice Hall. 3 y 0 3 2(x 5) or y 3 2(x 7) graph 5x 6y 30. 8x 2 4 6 40 10 Geometry Lesson 3-6 Geometry (0, 5) Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 3-6 All-In-One Answers Version B 41 L1 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 3-7 2 Finding Slopes for Perpendicular Lines Find theslope of a line Slopes of Parallel and Perpendicular Lines Lesson Objectives 1 Relate slope and parallel lines Relate slope and perpendicular lines 2 perpendicular to 5x 2y 1. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes To find the slope of the given line, rewrite the equation inslope-intercept form. Local Standards: ____________________________________ 5x 2y 1 2y 5 x 1 Key Concepts. 5 y 2 Slopes of Parallel Lines If the slopes of two distinct nonvertical lines are equal, the lines are parallel. Any two vertical lines are parallel. Slopes of Perpendicular Lines If the slopes oftwo lines have a product of 1, the lines are perpendicular. 5 ( ) Examples. © Pearson Education, Inc., publishing as Pearson Prentice Hall. slope-intercept All rights reserved. form. Write the equation x 5y 4 in slope-intercept form. x 5y 4 x 4 5y 5 4 5 Subtract 4 from each side. y y ⫺ Divide each side by 5. 15 x 4 5 The line x 5y 4 has slope ⫺1 5 . The line y 5x 4 has slope ⫺5 . The lines 42 are not parallel because their slopes are not equal Geometry Lesson 3-7 © Pearson Education, Inc., publishing as Pearson Prentice Hall. m x 5y 4 parallel? Explain. x © Pearson Education, Inc., publishing as PearsonPrentice Hall. Lesson Objectives 1 Construct parallel lines 2 Construct perpendicular lines 2 2 Multiply each side by 5 . 5 2 Simplify. 5 Quick Check. 1. Are the lines y 12x 5 and 2x 4y 9 parallel? Explain. Yes; Each line has slope 1 2 and the y-intercepts are different. 2. Find the slope of a line perpendicularto 5y x 10. m 5 . Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Lesson 3-8 . The product of the slopes of perpendicular lines is 1. m 1 ? 1 Determining Whether Lines are Parallel Are the lines y 5x 4and 1 5 2 m 1 2 Any horizontal line and vertical line are perpendicular. ⫺ Divide each side by 2. 2 Find the slope of a line perpendicular to 5x 2y 1. Let m be the slope of the perpendicular line. If two nonvertical lines are perpendicular, the product of their slopes is ⫺1. The equation y 5x 4 is in Subtract 5xfrom each side. 1 The line 5x 2y 1 has slope All rights reserved. All rights reserved. If two nonvertical lines are parallel, their slopes are equal. x Daily Notetaking Guide 43 Geometry Lesson 3-7 Name_____________________________________ Class____________________________Date________________ Example. Constructing Parallel and Perpendicular Lines R the 2 Perpendicular From a Point to a Line * Examine ) NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures construction. At what special point does RG meet line ᐉ? Point R is the samedistance from point E as it is from point F Local Standards: ____________________________________ / E F because the arc was made with one compass opening. G Point G is the same distance from point E as it is from point F Example. because both arcs were made with the same compass opening.1 Constructing ᐉm Examine the diagram at right. Explain m / H J All rights reserved. Step 1 With the compass point on point H, draw an arc that intersects the sides of ⬔H . Step 4 Use a straightedge to draw line m through the point you located and point N. Quick Check. 1. Use Example 1. Explain whylines ᐉ and m must be parallel. If corresponding angles are congruent, the lines are parallel by the Converse of Corresponding Angles Postulate. L1 ) ) Quick Check. midpoint of EF, perpendicular bisector * ) * ) of EF. * ) * ) Daily Notetaking Guide All-In-One Answers Version B L1 G © Pearson Education,Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Step 3 Put the compass point below point N where the arc * ) intersects HN . Open the compass to the length where the arc intersects line ᐉ . Keeping the same compass setting, put the compasspoint above point N where the * ) arc intersects HN . Draw an arc to locate a point. Geometry Lesson 3-8 * and that RG is the 2. Use a straightedge to draw EF . Construct FG so that FG ⬜ EF at point F. Step 2 With the same compass setting, put the compass point on point N . Draw an arc. 44 * Thismeans that RG intersects line ᐉ at the N 1 Use the method learned for constructing congruent angles. All rights reserved. how to construct ⬔1 congruent to ⬔H. Construct the angle. F E L1 Daily Notetaking Guide Geometry Lesson 3-8 Geometry 45 11 Geometry: All-In-One Answers Version B(continued) Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Congruent Figures List corresponding vertices inthe same order. If 䉭ABC 䉭CDE, then ⬔BAC NAEP 2005 Strand: Geometry Recognize congruent figures and their corresponding parts The diagram above shows ⬔BAC Topic: Transformation of Shapes and Preservation of Properties The statement 䉭ABC 䉭CDE Local Standards:____________________________________ Notice that BC Also, ⬔CBA Vocabulary and Key Concepts. two angles of another triangle, then the third angles B C E F ⬔ C ⬔ F Congruent polygons are polygons that have corresponding ⬔B ⬔ T 䉭 DEF GHI m⬔K Q T B ⬔ C ⬔J TJ AC C A QJ J Use theTriangle Angle-Sum Theorem and the definition of congruent polygons to find m⬔X. Triangle Angle-Sum Theorem m⬔Z m⬔M Corresponding angles of congruent triangles are congruent. m⬔Z 48 Substitute 48 for m⬔M. 115 180 Substitute. 180 Simplify. m⬔X 46 180 65 ⬔ACB . 䉭EDC . YS O KV lWYSO lMKV lSWY O lVMK 35 Corresponding angles of congruent triangles are congruent and have the same measure. 3. Can you conclude that 䉭JKL 䉭MNL? Justify your answer. N J L K M No. The corresponding sides are not necessarily equal. Subtract 115 from each side. Geometry Lesson 4-1 DailyNotetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Lesson Objective 1 Prove two triangles congruent using the SSS and SAS Postulates 47 Geometry Lesson 4-1Name_____________________________________ Class____________________________ Date ________________ 2 Using SAS AD > BC . What other information do you Triangle Congruence by SSS and SAS Lesson 4-2 Daily Notetaking Guide B A need to prove 䉭ADC 䉭BCD by SAS? It is giventhat AD > BC. Also, DC > CD by the NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties . Reflexive Property of Congruence ⬔ADC ⬵ ⬔BCD Therefore, if you know 䉭ADC 䉭BCD by SAS C D You now have two pairs of corresponding congruent sides. LocalStandards: ____________________________________ © Pearson Education, Inc., publishing as Pearson Prentice Hall. m⬔X . 2. It is given that 䉭WYS 䉭MKV. If m⬔Y 35, what is m⬔K? Explain. 2 Using Congruency 䉭XYZ 䉭KLM, m⬔Y 67, and m⬔M 48. Find m⬔X. 48 and ⬔BAC A . EC G F Listthe corresponding sides and angles in the same order. ⬔DEC WS O MV Angles: ⬔WSY ⬔MVK © Pearson Education, Inc., publishing as Pearson Prentice Hall. 䉭 1 Naming Congruent Parts 䉭ABC 䉭QTJ. List the congruent corresponding parts. 67 true. D 4 1. 䉭WYS 䉭MKV. List the congruentcorresponding parts. Use three letters for each angle. Sides: WY MK Examples. m⬔X is not DE , and AC 3 C 3 Quick Check. I E m⬔X m⬔Y m⬔Z , not ⬔DCE. are congruent. The correct way to state this is 䉭ABC H D sides congruent and corresponding angles congruent. © Pearson Education, Inc.,publishing as Pearson Prentice Hall. are congruent. All rights reserved. All rights reserved. A If two angles of one triangle are congruent to BC . Since all of the corresponding sides and angles are congruent, the triangles D Sides: AB > QT ⬔CDE ⬔DEC 4 B Using Theorem 4-1, you can conclude that⬔ECD Theorem 4–1 Angles: ⬔A ⬔Q DC , BA ⬔DCE All rights reserved. Lesson Objective 1 E 3 Finding Congruent Triangles Can you conclude that 䉭ABC 䉭CDE? Lesson 4-1 , you can prove . Key Concepts. Q P congruent. 䉭 GHF 䉭 F PQR Postulate 4-2: Side-Angle-Side (SAS) Postulate If twosides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles All rights reserved. Quick Check. H All rights reserved. Postulate 4-1: Side-Side-Side (SSS) Postulate If the three sides of one triangle are congruent to the three sides ofanother G triangle, then the two triangles are 1. Given: HF HJ, FG JK, H is the midpoint of GK. G D E C FDE A Examples. A 1 Proving Triangles Congruent Given: M is the midpoint of XY , AX > AY Prove: 䉭AMX 䉭AMY Write a paragraph proof. You are given that M is the midpoint of XY , and AX AY .Midpoint M implies that MX MY . AM > AM by the Reflexive Property of Congruence 䉭AMX 䉭AMY by the 48 12 Geometry Lesson 4-2 Geometry X , so SSS Postulate M Y © Pearson Education, Inc., publishing as Pearson Prentice Hall. 䉭 © Pearson Education, Inc., publishing as Pearson Prentice Hall.BCA H Statements are congruent. 䉭 J K R F B F Prove: 䉭FGH 䉭JKH Reasons 1. HF > HJ, FG > JK 1. Given 2. H is the midpoint of GK. 2. Given 3. GH > HK 3. Definition of Midpoint 4. 䉭FGH ⬵ 䉭JKH 4. SSS Postulate 2. What other information do you need to prove 䉭ABC 䉭CDA by SAS? A 8⬔DCA 艑 ⬔BAC 7 7 6 D B C . Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 4-2 All-In-One Answers Version B 49 L1 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________Date________________ Lesson 4-3 Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ Triangle Congruence by ASA and AAS Example. 2 Writing a Proof Write a two-column proof that uses AAS. NAEP 2005Strand: Geometry Prove two triangles congruent using the ASA Postulate and the AAS Theorem B A Given: ⬔B ⬔D, AB6CD Prove: 䉭ABC 䉭CDA Topic: Transformation of Shapes and Preservation of Properties Local Standards: ____________________________________ C D Key Concepts.Statements 1. ⬔B ⬔D, AB6CD P to two angles and the included side of another triangle, then 䉭 HGB 䉭 B K . G NKP N H 1. Given 2. lBAC O lDCA All rights reserved. If two angles and the included side of one triangle are congruent All rights reserved. Postulate 4-3: Angle-Side-Angle (ASA) Postulatethe two triangles are congruent Reasons alternate interior 2. If lines are parallel, then 2. angles are congruent. 3. AC > AC 3. Reflexive Property of Congruence 4. 䉭ABC 䉭CDA 4. AAS Theorem Theorem 4-2: Angle-Angle-Side (AAS) Theorem C If two angles and a nonincluded side of one triangle arecongruent Quick Check. to two angles and the corresponding nonincluded side of another CDM 䉭 G XGT All rights reserved. D O C and ⬔I is not congruent to ⬔C. Name the triangles that are congruent by the ASA Postulate. The diagram shows ⬔N ⬔A ⬔D and FN > CA > GD. F G T N A I If ⬔F ⬔C,then ⬔F ⬔C ⬔ G . Therefore, 䉭FNI 䉭CAT 䉭GDO by 2. In Example 2, explain how you could prove 䉭ABC 䉭CDA using ASA. ASA It is given that lB O lD. You know that lBAC O lDCA by the Alternate Interior Angles Theorem. By Theorem 4-1, lBCA O lDAC . By the Reflexive Property of Congruence,AC O AC . Therefore, kABC O kCDA by ASA. X Example. 1 Using ASA Suppose that ⬔F is congruent to ⬔C T D . Quick Check. 1. Using only the information in the diagram, can you conclude that 䉭INF is congruent to either of the other two triangles? Explain. © Pearson Education, Inc., publishing asPearson Prentice Hall. 䉭 M . © Pearson Education, Inc., publishing as Pearson Prentice Hall. the triangles are congruent triangle, then No; Only one angle and one side are shown to be congruent. At least one more congruent side or angle is necessary to prove congruence with SAS, ASA, or AAS. 50Geometry Lesson 4-3 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Geometry Lesson 4-3 51 Name_____________________________________ Class____________________________ Date________________ 2 Using Right Triangles According to legend, one of Napoleon’s followers used Lesson 4-4 Using Congruent Triangles: CPCTC Lesson Objective 1 Use triangle congruence and CPCTC to prove that parts of two triangles are congruent © Pearson Education, Inc., publishing asPearson Prentice Hall. Daily Notetaking Guide congruent triangles to estimate the width of a river. On the riverbank, the officer stood up straight and lowered the visor of his cap until the farthest thing he could see was the edge of the opposite bank. He then turned and noted the spot on his side of the riverthat was in line with his eye and the tip of his visor. NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties Local Standards: ____________________________________ G F Vocabulary. D CPCTC stands for: of C ongruent . Examples. E All rights reserved. P artsare C ongruent All rights reserved. C orresponding T riangles Given: ⬔DEG and ⬔DEF are right angles; ⬔EDG ⬔EDF. The officer then paced off the distance to this spot and declared that distance to be the width of the river! 1 Congruence Statements The diagram shows the frame of an umbrella. Whatcongruence statements besides ⬔3 ⬔4 can you prove from the diagram, in which SL > SR and ⬔1 ⬔2 are given? by the Reflexive Property of Congruence, 䉭RSC by SAS . ⬔3 ⬔4 because corresponding parts of congruent triangles are congruent. 12 When two triangles are congruent, you can formcongruence statements about three pairs of corresponding angles and three pairs of corresponding sides. List the congruence statements. S So the officer must stand ground must be Stretcher Shaft Sides: SL > SR Given SC SC Reflexive Property of Congruence CL CR Other congruence statementAngles: ⬔1 ⬔2 Given ⬔3 ⬔ 4 Corresponding Parts of Congruent Triangles lCLS O lCRS Other congruence statement right angles . ⬔DEG and ⬔DEF are the angles that the officer makes with the ground. R 6 L 5 © Pearson Education, Inc., publishing as Pearson Prentice Hall. SC and 䉭LSC Rib ©Pearson Education, Inc., publishing as Pearson Prentice Hall. SC The given states that ⬔DEG and ⬔DEF are What conditions must hold for that to be true? C 34 perpendicular level or flat to the ground, and the . Quick Check. 1. In Example 1, what can you say about ⬔5 and ⬔6? Explain. They arecongruent, because supplements of congruent angles are congruent. 2. Recall Example 2. About how wide was the river if the officer paced off 20 paces and each pace was about 212 feet long? 50 feet The congruence statements that remain to be proved are lCLS O lCRS and CL O CR. 52 L1 GeometryLesson 4-4 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 4-4 Geometry 53 13 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________Date________________ Lesson 4-5 Example. Isosceles and Equilateral Triangles Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ Using the Isosceles Triangle Theorems Explain why 䉭ABC is isosceles. NAEP2005 Strand: Geometry Use and apply properties of isosceles triangles Topic: Relationships Among Geometric Figures ⬔ABC and ⬔XAB are Local Standards: ____________________________________ alternate interior X A angles formed * ) ← → ← → ← → ← → by XA , BC , and the transversal AB. Because XA 6 BC , ⬔ABC ⬔ Vocabulary and Key Concepts. XAB . B C The diagram shows that ⬔XAB ⬔ACB. By Theorem 4-3: Isosceles Triangle Theorem ⬔ A ⬔ B B A C All rights reserved. All rights reserved. the angles opposite those sides are congruent. . to AC . By the definition of an isoscelestriangle, 䉭ABC is isosceles. a. In the figure, suppose m⬔ACB 55. Find m⬔CBA and m⬔BAC. the sides opposite those sides are congruent. conclude that AB > ACB Quick Check. If two angles of a triangle are congruent, then AC the Converse of the Isosceles Triangle Theorem You can use If two sidesof a triangle are congruent, then Theorem 4-4: Converse of Isosceles Triangle Theorem , ⬔ABC ⬔ the Transitive Property of Congruence C Theorem 4-5 m⬔CBA 55; m⬔BAC 70 B A BC C and CD A bisects AB. The legs of an isosceles triangle D The are the congruent sides. B vertex angle of anisosceles triangle is formed by Vertex the two congruent sides (legs). Angle The base angles of an isosceles The of an base and the legs. Base non-congruent, side. 54 triangle are formed by the base Legs isosceles triangle is the third, or Base Angles Geometry Lesson 4-5 Daily Notetaking Guide L1 b. Inthe figure, can you deduce that 䉭ABX is isosceles? No; neither lAXB nor lXBA can be shown to be congruent to lA. Daily Notetaking Guide L1 Name_____________________________________ Class____________________________ Date________________ All rights reserved. AB Geometry Lesson4-5 Name_____________________________________ Class____________________________ Date ________________ 2 Two-Column Proof—Using the HL Theorem Lesson 4-6 Congruence in Right Triangles Lesson Objective 1 Prove triangles congruent using the HL Theorem 55 C Given: ⬔ABCand ⬔DCB are right angles, AC > DB Prove: 䉭ABC 䉭DCB NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties Statements Reasons 1. ⬔ABC and ⬔DCB are Local Standards: ____________________________________ D E B © Pearson Education, Inc.,publishing as Pearson Prentice Hall. CD ' © Pearson Education, Inc., publishing as Pearson Prentice Hall. of the base. perpendicular bisector © Pearson Education, Inc., publishing as Pearson Prentice Hall. The bisector of the vertex angle of an isosceles triangle is the A 1. Given right angles 2. 䉭ABCand 䉭 Vocabulary and Key Concepts. DCB 2. Definition of Right Triangle are right triangles All rights reserved. the triangles are congruent. All rights reserved. 3. AC > Theorem 4-6: Hypotenuse-Leg (HL) Theorem If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and leg ofanother right triangle, then 3. Given DB BC O CB 4. 5. 䉭ABC 䉭 4. Reflexive Property of Congruence DCB 5. HL The hypotenuse of a right triangle is its longest side, or the side opposite the right angle. Quick Check. The of a right triangle are the two shortest sides, or the sides that are not legs 1. R Lopposite the right angle. 3 S O 3 Examples. 1 Proving Triangles Congruent One student wrote “䉭CPA 䉭MPA A by the HL Theorem” for the diagram. Is the student correct? Explain. The diagram shows the following congruent parts. CA > MA ⬔CPA lMPA PA PA C hypotenuse Since AC is the of righttriangle CPA, and MA is the leg the 56 14 hypotenuse M leg 5 3 M 5 5 Q P N T Which two triangles are congruent by the HL Theorem? Write a correct congruence statement. 䉭LMN 艑 䉭OQP 2. You know that two legs of one right triangle are congruent to two legs of another right triangle. Explain how toprove the triangles are congruent. Since all right angles are congruent, the triangles are congruent by SAS. and PA is a of right triangle MPA, the triangles are congruent by HL Theorem The student and PA is a P © Pearson Education, Inc., publishing as Pearson Prentice Hall. Legs © Pearson Education,Inc., publishing as Pearson Prentice Hall. Hypotenuse is . correct. Geometry Lesson 4-6 Geometry Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 4-6 All-In-One Answers Version B 57 L1 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ Lesson 4-7 Lesson Objectives 1 Quick Check. 1. The diagram shows triangles from the scaffolding that workers used when they repaired and cleaned the Statue of Liberty. a. Name thecommon side in 䉭ACD and 䉭BCD. NAEP 2005 Strand: Geometry Identify congruent overlapping triangles Prove two triangles congruent by first proving two other triangles congruent 2 Name_____________________________________ Class____________________________ Date________________ Using Corresponding Parts of Congruent Triangles Topic: Relationships Among Geometric Figures Local Standards: ____________________________________ E E D , respectively. FG 2 Using Two Pairs of Triangles Write a paragraph proof. X All rights reserved. and HG All rightsreserved. EG These parts are F H are shared by 䉭DFG and 䉭EHG. Parts of sides DG and Answers may vary. Sample: 䉭ABD and 䉭CBD; BD G 䉭DFG and 䉭EHG share. Identify the overlapping triangles. 2. Write a two-column proof. Given: PS > RS, ⬔PSQ ⬔RSQ Prove: 䉭QPT 䉭QRT Plan: 䉭XPW䉭YPZ by AAS if ⬔WXZ SAS S lZYW . These angles P Proof: You are given XW > YZ. Because ⬔XWZ and ⬔YZW are Reflexive Property of Congruence 䉭XPW 䉭YPZ by . Therefore, 䉭XWZ 䉭YZW by by CPCTC, and ⬔XPW ⬔ ZYW vertical angles because W , ⬔XWZ ⬔YZW. WZ > ZW by the rightangles SAS. ⬔WXZ ⬔ Z YPZ are congruent. Therefore, AAS . Geometry Lesson 4-7 Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. . © Pearson Education, Inc., publishing as Pearson Prentice Hall. congruent by R T Statements if 䉭XWZ 䉭YZW. These trianglesare CPCTC are congruent by Q P Y Given: XW > YZ, ⬔XWZ and ⬔YZW are right angles. Prove: 䉭XPW 䉭YPZ All rights reserved. D b. Name another pair of triangles that share a common side. Name the common side. 1 Identifying Common Parts Name the parts of the sides that L1 L1Name_____________________________________ Class____________________________ Date________________ Reasons 1. PS O RS, lPSQ O lRSQ 1. Given 2. QS > QS 2. Reflexive Property of Congruence 3. 䉭PSQ 䉭RSQ 3. SAS 4. PQ > RQ 4. CPCTC 5. ⬔PQT lRQT 5. CPCTC 6. QT > QT 6.Reflexive Property of Congruence 7. 䉭QPT 䉭QRT 7. SAS Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ Midsegments of Triangles Lesson Objective 1 Use properties of midsegments to solve problems 59Geometry Lesson 4-7 2 Identifying Parallel Segments Find m⬔AMN and m⬔ANM. MN and BC Lesson 5-1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. C F B CD Examples. 58 A are cut by transversal AB, so ⬔AMN and ⬔B are NAEP 2005 Strand: Geometry Topics: RelationshipsAmong Geometric Figures angles. MN6BC by the Triangle Midsegment ⬔AMN ⬔B by the Corresponding Angles Local Standards: ____________________________________ m⬔AMN 75 Theorem, so Postulate. N because congruent angles have the same measure. In 䉭AMN, AM AN , so m⬔ANMVocabulary and Key Concepts. A corresponding Isosceles Triangle m⬔AMN M by the Theorem. m⬔ANM 75 C 75 B by substitution. parallel is half to the third side, and is All rights reserved. If a segment joins the midpoints of two sides of a triangle, then the segment its length. A midsegment of a triangleis a segment connecting the midpoints of two sides. A coordinate proof All rights reserved. Theorem 5-1: Triangle Midsegment Theorem Quick Check. 1. AB 10 and CD 18. Find EB, BC, and AC. A E B D is a form of proof in which coordinate geometry and algebra are used to prove a theorem. C EB ⫽ 9;BC ⫽ 10; AC ⫽ 20 Examples. 1 Finding Lengths In 䉭XYZ, M, N, and P are midpoints. NP MN MP 60 M 22 P Definition of perimeter 24 NP 24 NP 22 46 60 60 NP Substitute 22 24 for MN and Y for MP. N Z Simplify. 14 46 Subtract from each side. Use the Triangle Midsegment Theorem to find YZ. MP 60L1 1 YZ Triangle Midsegment Theorem 2 1 22 44 YZ 2 YZ Substitute 22 Multiply each side by Geometry Lesson 5-1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Because the perimeter of 䉭MNP is 60, you can find NP. © Pearson Education, Inc., publishing as Pearson Prentice Hall.The perimeter of 䉭MNP is 60. Find NP and YZ. 2. Critical Thinking Find m⬔VUZ. Justify your answer. X 65 Y U V Z 65; UV6XY so ⬔VUZ and ⬔YXZ are corresponding and congruent. for MP. 2 . Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 5-1Geometry 61 15 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 5-2 Example. Bisectors in Triangles Lesson Objective 1Name_____________________________________ Class____________________________ Date ________________ Using the Angle Bisector Theorem Find x, FB, and FD in the diagram at right. NAEP 2005 Strand: Geometry Use properties of perpendicular bisectors and angle bisectors Topics:Relationships Among Geometric Figures FD Local Standards: ____________________________________ FB 7x 37 2x 5 Vocabulary and Key Concepts. 7x 2x equidistant from the endpoints of the segment. Theorem 5-3: Converse of the Perpendicular Bisector Theorem If a point is equidistant from theendpoints of a segment, then it is x All rights reserved. All rights reserved. If a point is on the perpendicular bisector of a segment, then it is on the perpendicular bisector of the segment. Substitute. 42 D 2x 5 A Add 37 to each side. 42 5x Theorem 5-2: Perpendicular Bisector Theorem C B Angle BisectorTheorem 8.4 Substitute. FD 7( 8.4 ) 37 21.8 Substitute. Quick Check. ) ) D 10; 10 2x If a point is on the bisector of an angle, then it is K angle, then it is on the angle bisector. The distance from a point to a line is the length of the perpendicular segment from the point to the line. * ) D is 3 in. from AB and * )AC . C 0 B ) (x 20) H b. What can you conclude about EK ? ) EK is the angle bisector of lDEH. All rights reserved. If a point in the interior of an angle is equidistant from the sides of the E C 10 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing asPearson Prentice Hall. equidistant from the sides of the angle. Theorem 5-5: Converse of the Angle Bisector Theorem 1 2 3 4 5 6 D E Divide each side by 5. a. According to the diagram, how far is K from EH ? From ED ? A F Subtract 2x from each side. FB 2( 8.4 ) 5 21.8 Theorem 5-4: Angle BisectorTheorem 7x 37 c. Find the value of x. 20 d. Find m⬔DEH. 80 Geometry Lesson 5-2 Daily Notetaking Guide L1 Name_____________________________________ Class____________________________ Date________________ Lesson Objective 1 Identify properties of perpendicular bisectors andangle bisectors 2 Identify properties of medians and altitudes of a triangle 2 Finding Lengths of Medians M is the centroid of 䉭WOR, and WM 16. W Find WX. NAEP 2005 Strand: Geometry Topics: Relationships Among Geometric Figures The is the point of concurrency of the medians of centroid Z Y atriangle. M Local Standards: ____________________________________ The medians of a triangle are concurrent at a point that is O R X the distance from each vertex to the midpoint two-thirds of the opposite side. (Theorem 5-8) Vocabulary. of 䉭WOR, WM centroid Because M is the is the point atwhich concurrent lines intersect. point of concurrency A circle is circumscribed about a polygon when the vertices Circumscribed of the polygon are on the circle. circumcenter S of a triangle is the point of concurrency of the perpendicular bisectors of a triangle. Median A median of a triangle is a segmentthat circle QC SC RC All rights reserved. The All rights reserved. Concurrent lines are three or more lines that meet in one point. The 63 Geometry Lesson 5-2 Name_____________________________________ Class____________________________ Date ________________ Concurrent Lines,Medians, and Altitudes Lesson 5-3 Daily Notetaking Guide L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 62 circumcenter Examples. 1 Finding the Circumcenter Find the center of the circle that (1, 1) and Y has , XY lies on the vertical line x 1 . The perpendicular bisector of XY is thehorizontal line that 7 passes through (1, 1 2 ) or (1, 4) , so the equation of the perpendicular bisector of XY is y 4 . (1, 1) Because X has coordinates coordinates (5, 1) and Z has y 8Y 6 (3, 4) 4 2 X O Z 2 4 6 x , XZ lies on the horizontal line y 1 . The perpendicular bisector of XZ is the vertical line thatpasses through (1 2 5, 1) or (3, 1) © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. side. (1, 7) Theorem WX Substitute 2WX 3 . 5-8 2 16 24 WX 16 for WM. 3 Multiply each side by 3 2 . y Quick Check. triangle and themidpoint of the opposite coordinates WX R Q circumscribes 䉭XYZ. 2 3 C has as its endpoints a vertex of the Because X has coordinates WM 1. a. Find the center of the circle that you can circumscribe about the triangle with vertices (0, 0), (8, 0), and (0, 6). 5 (⫺4, 3) 5 5x 5 b. Critical Thinking In Example1, explain why it is not necessary to find the third perpendicular bisector. Theorem 5-6: All the perpendicular bisectors of the sides of a triangle are concurrent. 2. Using the diagram in Example 2, find MX. Check that WM MX WX. , so the equation of the perpendicular bisector of XZ is x 3 . 8 Draw the linesy 4 and x 3 . They intersect at the point (3, 4) . This point is the center of the circle that circumscribes 䉭XYZ. 64 16 Geometry Lesson 5-3 Geometry Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 5-3 All-In-One Answers Version B 65 L1 Geometry: All-In-One Answers Version B(continued) Name_____________________________________ Class____________________________ Date ________________ Lesson 5-4 Lesson Objectives 1 2 Writing the Inverse and Contrapositive Write the inverse and contrapositive of the conditional statement “If 䉭ABC is equilateral, then it isisosceles.” To write the inverse of a conditional, negate both the hypothesis and the conclusion. NAEP 2005 Strand: Geometry Topics: Mathematical Reasoning Write the negation of a statement and the inverse and contrapositive of a conditional statement 2Name_____________________________________ Class____________________________ Date ________________ Inverses, Contrapositives, and Indirect Reasoning Hypothesis Inverse: If an angle is a straight angle, then its measure is 180. Symbolic Form You Read It pSq p, If q. then Negation (ofp) An angle is not a straight angle. p Not Inverse If an angle is not a straight angle, then its measure is not 180. p S q If not p, then not q. Contrapositive If an angle’s measure is not 180, then it is not a straight angle. q S p If not q, then not p. The contrapositive of a conditional switches the hypothesis andthe conclusion and negates both. contrapositive always have the same truth value. Equivalent statements have the same truth value. Examples. 1 Writing the Negation of a Statement Write the negation of “ABCD is not a convex polygon.” opposite The negation of a statement has the . HypothesisConclusion T T Switch and negate both. , then it is not isosceles . not equilateral Quick Check. © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. conclusion. and its T not isosceles then it is If 䉭ABC is equilateral, then it is isosceles. Conditional: Contrapositive: If 䉭ABCis of a conditional statement negates both the hypothesis and the conditional , p. The negation of a statement has the opposite truth value from that of the original statement. inverse not equilateral To write the contrapositive of a conditional, switch the hypothesis and conclusion, then negate both. All rightsreserved. Example Conditional © Pearson Education, Inc., publishing as Pearson Prentice Hall. Statement All rights reserved. Negation, Inverse, and Contrapositive Statements If 䉭ABC is then it is isosceles. Negate both. T Vocabulary and Key Concepts. A T If 䉭ABC is equilateral, Conditional: Useindirect reasoning The Conclusion T Local Standards: ____________________________________ 1. Write the negation of each statement. a. m⬔XYZ 70. The measure of ⬔XYZ is not more than 70. b. Today is not Tuesday. Today is Tuesday. 2. Write (a) the inverse and (b) the contrapositive of MayaAngelou’s statement “If you don’t stand for something, you’ll fall for anything.” a. b. If you stand for something, you won’t fall for anything. If you won’t fall for anything, then you stand for something. truth value. The negation of is not in the original statement removes the word not . The negation of “ABCD isnot a convex polygon” is “ABCD is a convex polygon.” 66 Geometry Lesson 5-4 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Geometry Lesson 5-4 67Name_____________________________________ Class____________________________ Date ________________ 2 Using the Triangle Inequality Theorem Can a triangle have sides with the given lengths? Lesson 5-5 Inequalities in Triangles Lesson Objectives 1 Use inequalities involving angles oftriangles 2 Use inequalities involving sides of triangles © Pearson Education, Inc., publishing as Pearson Prentice Hall. Daily Notetaking Guide Explain. a. 2 cm, 2 cm, 4 cm NAEP 2005 Strand: Geometry Topics: Relationships Among Geometric Figures Triangle Inequality According to the Theorem, thesum of the lengths of any two sides of a triangle is Local Standards: ____________________________________ greater than the length of the third side. 22 ⬐ 4 Key Concepts. The sum of 2 and 2 No , a triangle cannot is not greater than 4. have sides with these lengths. b. 8 in., 15 in., 12 in. m⬔1 m⬔2and 3 angle of a triangle is greater exterior than the measure of each of its remote angles. interior 2 m⬔3 m⬔1 1 All rights reserved. The measure of an All rights reserved. Corollary to the Triangle Exterior Angle Theorem 8 15 ⬎ 12 Yes , a triangle 8 12 ⬎ 15 can 15 12 ⬎ 8 have sides with these lengths.Quick Check. 1. List the angles of 䉭ABC in order from smallest to largest. Theorem 5-10 If two sides of a triangle are not congruent, then A ⬔A, ⬔C, ⬔B Y 27 ft C the larger angle lies opposite the longer side. X If XZ XY, then m⬔Y ⬎ m⬔Z. 21 ft Z 18 ft greater than the length of the third side. LM MN LNMN LN LM LN LM MN M N Examples. 1 Applying Theorem 5-10 In 䉭RGY, RG 14, GY 12, and RY 20. G List the angles from largest to smallest. No two sides of 䉭RGY are congruent, so the largest angle lies opposite the longest side. Find the angle opposite each side. The longest side is 20 12 Y 20 .The opposite angle, ⬔G , is the largest. The shortest side is 12 14 R © Pearson Education, Inc., publishing as Pearson Prentice Hall. L The sum of the lengths of any two sides of a triangle is © Pearson Education, Inc., publishing as Pearson Prentice Hall. Theorem 5-12: Triangle Inequality Theorem B 2.Can a triangle have sides with the given lengths? Explain. a. 2 m, 7 m, and 9 m no; 2 7 is not greater than 9 b. 4 yd, 6 yd, and 9 yd yes; 4 6 ⬎ 9, 6 9 ⬎ 4, and 4 9 ⬎ 6 . The opposite angle, ⬔R , is the smallest. From largest to smallest, the angles are ⬔G , ⬔Y , ⬔R . 68 L1 Geometry Lesson 5-5 DailyNotetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 5-5 Geometry 69 17 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 6-1Example. Classifying Quadrilaterals Classifying by Coordinate Methods Determine the most precise name for the quadrilateral with vertices Q(4, 4), B(2, 9), H(8, 9), and A(10, 4). Graph quadrilateral QBHA. NAEP 2005 Strand: Geometry Define and classify special types of quadrilaterals Topic:Relationships Among Geometric Figures Local Standards: ____________________________________ Find the slope of each side. 94 2 (4) 99 Slope of BH 8 2 y B Slope of QB H Key Concepts. 8 Special Quadrilaterals Parallelogram Rhombus Isosceles Trapezoid All rights reserved. All rights reserved.Rectangle Q Trapezoid Kite ( 6 Quadrilateral Quadrilateral of irs es 2 pairs of pa 1 pair of id No llel s parallel sides parallel sides ra pa Square A Slope of HA 2 4 6 8 equal BH is parallel to QA because their slopes are square A is a parallelogram with congruent sides ___ four________________________________ and four right angles ___ ___________________ . rectangle A is a parallelogram with right angles four______________________ __________________________. isosceles trapezoid An _____________________ ________________________ is a trapezoid whose nonparallel sidesare congruent. Geometry Lesson 6-1 Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. kite A is a quadrilateral with two pairs adjacent sides of ______________________ opposite congruent and no________ sides ________________________ congruent. trapezoid A is a quadrilateral with exactly one pair of ________________________ parallel sides ________________________ ________________________ . L1 not equal "29 "29 (2 (4))2 (9 4 )2 4 25 (10 8)2 ( 4 9 )2 4 25 BH (8 (2))2 ( 9 9 )2 1000 2 2 QA (4 10 ) ( 4 4 ) 196 0 Because QB 0 HA 10 14 an isosceles trapezoid , QBHA is . Quick Check. y a. Graph quadrilateral ABCD with vertices A(3, 3), B(2, 4), C(3, 1), and D(2, 2). b. Classify ABCD in as many ways as possible. B A x ABCD is a quadrilateral because it has four sides. It is aparallelogram because both pairs of opposite sides are parallel. It is a rhombus because all four sides are congruent. It is a rectangle because it has four right angles and its opposite sides are congruent. It is a square because it has four right angles and its sides are all congruent. C D Daily NotetakingGuide 2x 5 L x A 7y 16 5y Local Standards: ____________________________________ M N x 7y 16 A Opposite angles of a parallelogram are congruent . C D R 14y 14y 32 5 5y 27 bisect U each other. 2x 120 Add x to each side. 60 Divide each side by 2. 15 75 Substitute 60 75 180 105 D C for x.Consecutive angles of a parallelogram are supplementary m⬔A (135 x) Subtract 15 from each side. x m⬔A m⬔B 180 (x 15) . Substitute Subtract 75 75 . for m⬔B. from each side. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 135 © Pearson Education, Inc., publishing as PearsonPrentice Hall. congruent B A Subtract 14y Divide each side by from each side. 9 . x7 3 Substitute 3 for y in the first equation to solve for x. x 5 Simplify. ( ) 16 T Opposite angles of a parallelogram are Simplify. 9y 3 y Examples. 1 Using Algebra Find the value of x in ⵥABCD. Then find m⬔A. 15 each other.Distribute. 5y 27 S Theorem 6-3 x 15 135 x bisect Substitute 7y 16 for x in the second equation to solve for y. 2(7y 16) 5 5y All rights reserved. Theorem 6-2 All rights reserved. B sides of a parallelogram are congruent. The diagonals of a parallelogram The diagonals of a parallelogram 2x 5 5y Theorem 6-1 Opposite So x 5 and y 3 . Quick Check. 1. Find the value of y in ⵥEFGH. Then find m⬔E, m⬔F, m⬔G, and m⬔H. F y 5 11; mlE 5 70, mlF 5 110, mlG 5 70, and mlH 5 110 (6y 4) E 2. Find the values of a and b. (3y 37) X a a ⫽ 16, b ⫽ 14 b 0 1 18 Geometry Daily Notetaking Guide L1 L1 Daily NotetakingGuide G H Y 8 2a b 2 W Geometry Lesson 6-2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures Vocabulary and Key Concepts. 72 71 Geometry Lesson 6-1 Name_____________________________________Class____________________________ Date ________________ K m⬔A 25 2 . HA Properties of Parallelograms Lesson Objectives 1 Use relationships among sides and among angles of parallelograms 2 Use relationships involving diagonals of parallelograms and transversals 60 2 Using Algebra Findthe values of x and y in ⵥKLMN. Lesson 6-2 m⬔B 10 0 . QB L1 Name_____________________________________ Class____________________________ Date________________ 2x 4 5 2 trapezoid One pair of opposite sides is parallel, so QBHA is a . Next, use the distance formula to see whetherany pairs of sides are congruent. parallelogram A is a quadrilateral with both pairs of opposite parallel sides ___________________ ________________________ . 70 4 4 Slope of QA x HA is not parallel to QB because their slopes are rhombus A is a parallelogram with congruent sidesfour_____________________ ______________________. 10 8 2 4 2 O ) 4 9 All rights reserved. Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ Z Geometry Lesson 6-2 All-In-One Answers Version B 73 L1Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 6-3 Lesson Objective 1 Name_____________________________________ Class____________________________ Date________________ Proving That a Quadrilateral Is a Parallelogram 2 Is the Quadrilateral a Parallelogram? Can you prove the quadrilateral is a parallelogram from what is given? Explain. NAEP 2005 Strand: Geometry Determine whether a quadrilateral is a parallelogram Given: m⬔E m⬔G 75°, m⬔F105° Prove: EFGH is a parallelogram. Topic: Geometry Local Standards: ____________________________________ ( the quadrilateral is a parallelogram. of a quadrilateral are congruent, then the quadrilateral is a parallelogram. bisect congruent and parallel All rights reserved. Examples. A B 8x 12 forwhich ABCD must be a parallelogram. y9 2y 80 G 10x 24 C D Diagonals of parallelograms bisect each other. 10x 24 8x 12 2x If x 74 24 12 Collect the variables on one side. 2x 36 Solve. x 18 and y 18 2y 80 y9 1. Find the values of a and c for which PQRS must be a parallelogram. Given: PQ > SR, PQ6SR Prove: PQRS is a parallelogram. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Q Yes; a pair of opposite sides are parallel and congruent (Theorem 6-8). R x P L1 S Daily Notetaking Guide L1 Geometry Lesson 6-3 Examples. Special Parallelograms 1 Finding Angle Measures Findthe measures of the B numbered angles in the rhombus. Theorem 6-9 states that each diagonal of a rhombus bisects two angles of the rhombus, so m⬔1 78. Local Standards: ____________________________________ 78 3 the diagonals of the rhombus are perpendicular so m⬔2 90 . Because the fourangles formed by the diagonals A Theorem 6-10 B perpendicular BD A Rectangles Theorem 6-11 The diagonals of a rectangle are congruent the complementary . . Finally, because BC DC, 78 . 2 Finding Diagonal Length One diagonal of a rectangle has length 8x 2. The other diagonal has length 5x 11.Find the length of each diagonal. C By Theorem 6-11, the diagonals of a rectangle are 5x 11 D A D B C . BD 12 allows you to conclude that ⬔ 1 ⬔4. Isosceles Triangle Theorem So m⬔4 . Parallelograms Theorem 6-12 If one diagonal of a parallelogram bisects two angles of the parallelogram, theparallelogram is a rhombus. Theorem 6-13 If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. © Pearson Education, Inc., publishing as Pearson Prentice Hall. The diagonals of a rhombus are D Because m⬔ABD 78, m⬔3 90 78 All rights reserved. , so ⬔3 ⬔4 4 ©Pearson Education, Inc., publishing as Pearson Prentice Hall. BCD 1 2 All rights reserved. Each diagonal of a rhombus bisects two angles of the rhombus. AC bisects ⬔BAD, so ⬔ 1 ⬔ 2 C 4 D , all must have measure 90, ⬔3 and ⬔ABD must be 3 C 1 2 A Theorem 6-10 states that B 75

Name_____________________________________ Class____________________________ Date ________________ NAEP 2005 Strand: Geometry Topic: Geometry Rhombuses Theorem 6-9 c1 S 2. Can you prove the quadrilateral is a parallelogram? Explain. Key Concepts. AC 3c 3 89 DailyNotetaking Guide Lesson Objectives 1 Use properties of diagonals of rhombuses and rectangles 2 Determine whether a parallelogram is a rhombus or a rectangle ' a x Lesson 6-4 AC R (a 40) P Name_____________________________________ Class____________________________Date________________ bisects ⬔ Q a 70, c 2 89 , then ABCD is a parallelogram. Geometry Lesson 6-3 AC . Theorem 6-6 y 80 9 y H Quick Check. , then the quadrilateral is a parallelogram. If the diagonals of quadrilateral ABCD bisect each other, Theorem 6-7 then ABCD is a parallelogram by . Writeand solve two equations to find values of x and y for which the diagonals bisect each other. 75 E Because both pairs of opposite angles are congruent, the quadrilateral is a parallelogram by 1 Finding Values for Parallelograms Find values for x and y G congruent, then the quadrilateral is a parallelogram.each other, Theorem 6-8 If one pair of opposite sides of a quadrilateral is both 75 Theorem 6-6 states If both pairs of opposite angles of a quadrilateral are © Pearson Education, Inc., publishing as Pearson Prentice Hall. Theorem 6-7 If the diagonals of a quadrilateral then the quadrilateral is aparallelogram. © Pearson Education, Inc., publishing as Pearson Prentice Hall. opposite angles 105 ) If x represents the measure of the unmarked angle, 360 105 x 75 105 75 , so x = . All rights reserved. All rights reserved. Theorem 6-5 If both pairs of opposite sides of a quadrilateral are congruent, thenTheorem 6-6 If both pairs of F The sum of the measures of the angles of a polygon is (n 2)180 where n represents the number of sides, so the sum of the measures of the angles of a 360 quadrilateral is 4 2 180 . Key Concepts. 8x 2 11 3x 9 3x 2 Subtract . 5x from each side. Subtract 2 from each side. 3 x( )2 5x 11 5( 3 ) 11 8x 2 8 3 congruent Diagonals of a rectangle are congruent. Divide each side by 3 . 26 Substitute. 26 The length of each diagonal is Substitute. 26 . Quick Check. 1. Find the measures of the numbered angles in the rhombus. 50 1 m⬔1 90, m⬔2 50, m⬔3 50, m⬔4 40 4 2. Find thelength of the diagonals of rectangle GFED if FD 5y 9 and GE = y 5. 2 3 F E G D 81 2 Theorem 6-14 If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. 76 L1 Geometry Lesson 6-4 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking GuideGeometry Lesson 6-4 Geometry 77 19 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ 2 Finding Angle Measures in Kites Find m⬔1, m⬔2, and m⬔3 in the kite. Lesson 6-5Trapezoids and Kites Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ S 1 NAEP 2005 Strand: Geometry Verify and use properties of trapezoids and kites Topic: Relationships Among Geometric Figures D R 3Local Standards: ____________________________________ U Vocabulary and Key Concepts. m⬔2 90 RU . Theorem 6-16 diagonals of an isosceles trapezoid are congruent. m⬔1 All rights reserved. congruent The base angles of an isosceles trapezoid are All rights reserved. Trapezoids Theorem 6-15 The T 2 72 Diagonals of a kite are Definition of a kite 72 Isosceles Triangle Theorem m⬔3 m⬔RDU 72 180 m⬔RDU m⬔3 m⬔3 90 72 180 90 162 180 m⬔3 The base angles of a trapezoid are two angles that share a base of the trapezoid. perpendicular RS 18 . Triangle Angle-Sum TheoremDiagonals of a kite are perpendicular. Substitute. Simplify. Subtract 162 from each side. Quick Check. angles angles 1. In the isosceles trapezoid, m⬔S 70. Find m⬔P, m⬔Q, and m⬔R. Leg perpendicular . Examples. 1 Finding Angle Measures in Trapezoids WXYZ is X 156 an isosceles trapezoid, andm⬔X 156. Find m⬔Y, m⬔Z, and m⬔W. Y Z W Two angles that share a leg of a trapezoid 180 are 156 m⬔W 180 m⬔W . supplementary S 70 R 2. Find m⬔1, m⬔2, and m⬔3 in the kite. m⬔1 90, m⬔2 46, and m⬔3 44 46 2 1 3 Substitute. 24 156 Subtract from each side. Because the base angles of anisosceles trapezoid are m⬔Y m⬔X 156 and m⬔Z 78 Q m⬔W 24 , congruent . Geometry Lesson 6-5 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Lesson Objective 1 Name coordinates of specialfigures by using their properties 79 Geometry Lesson 6-5 Name_____________________________________ Class____________________________ Date________________ Placing Figures in the Coordinate Plane Lesson 6-6 Daily Notetaking Guide Example. 2 Naming Coordinates Use theproperties of parallelogram NAEP 2005 Strand: Geometry Topic: Position and Direction y OCBA to find the missing coordinates. Do not use any new variables. (0, 0) The vertex O is the origin with coordinates O . Local Standards: ____________________________________ (p, q) A B © PearsonEducation, Inc., publishing as Pearson Prentice Hall. m⬔X m⬔W © Pearson Education, Inc., publishing as Pearson Prentice Hall. Kites Theorem 6-17 The diagonals of a kite are P 110, 110, 70 All rights reserved. Base Base © Pearson Education, Inc., publishing as Pearson Prentice Hall. Leg Becausepoint A is p units to the left of point O, point B Example. is also p units to the left of point C , because OCBA is a If both pairs of opposite sides of a quadrilateral are congruent T(a, b) x . Theorem 6-6 You can prove that TWVU is a parallelogram by showing that TW VU and WV U(e, 0) O E TU . Use thedistance formula. ( ) ( ) ) ( ) 2 2 ce e) ( d 0 ) c2 d2 2 2 ce ) ( bd d ) ac 2 2 a e ) ( b 0 ) (a e) 2 b 2 2 bd Because TW VU and WV TU, TWVU is a b 2 c2 parallelogram Geometry O x (0, 0) and B (p x, q) . y R(b, c) Q(?, ?) Q(s b, c) O (a e) 2 b 2 . Midpoint of TV 5 aa 1 c 1 e , b 1 d b midpoint of UW . Thus, the2 2 diagonals bisect each other, and TWVU is a parallelogram. 20 The missing coordinates are O 2. Use the properties of parallelogram OPQR to find the missing coordinates. Do not use any new variables. 1. Use the diagram above. Use a different method: Show that TWVU is a parallelogram by findingthe midpoints of the diagonals. Geometry Lesson 6-6 C (x, 0) . So point B has the same second coordinate, q , as point A. d2 Quick Check. 80 horizontal Daily Notetaking Guide L1 P(s, 0) x © Pearson Education, Inc., publishing as Pearson Prentice Hall. a © Pearson Education, Inc., publishing asPearson Prentice Hall. ( VU ( WV ( TU ( ac . Quick Check. ( ) Use the coordinates T (a, b), W a c , b d , V c e , d , and U e, 0 . TW p x Because AB 6 CO and CO is horizontal, AB is also V(c e, d) , then the quadrilateral is a parallelogram by parallelogram. So the first coordinate of point B is W(a c, b d) CAll rights reserved. A y proving pairs of opposite sides congruent. All rights reserved. 1 Proving Congruency Show that TWVU is a parallelogram by L1 Daily Notetaking Guide Geometry Lesson 6-6 All-In-One Answers Version B 81 L1 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 6-7 If the diagonals of a parallelogram are Proofs UsingCoordinate Geometry Lesson Objective 1 parallelogram is a NAEP 2005 Strand: Geometry Prove theorems using figures in the coordinate plane Topics: Position and Direction; Mathematical Reasoning To show that XYZW is a rectangle, find the lengths of its diagonals, and then Local Standards:____________________________________ compare them to show that they are XZ Vocabulary and Key Concepts. Theorem 6-18: Trapezoid Midsegment Theorem All rights reserved. to the bases. (2) The length of the midsegment of a trapezoid is half the sum of the . All rights reserved. parallel (1) Themidsegment of a trapezoid is lengths of the bases The midsegment of a trapezoid is the segment that joins the midpoints of the nonparallel R RA ) All rights reserved. y D(0, 2b) W(a, b) A(2a, 0) x © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing asPearson Prentice Hall. TP Example. C(2a, 0) . )) 2 ⫺2a 2 4a2 4b2 (a a )2 (b ( ( ⫺2a )2 ( ⫺2b )2 4a2 4b2 YW b ))2 rectangle . Quick Check. P 1 RA , and MN 2 ( Using Coordinate Geometry Use coordinate geometry to prove that the quadrilateral formed by connecting the midpoints of rhombus ABCD is arectangle. Draw quadrilateral XYZW by connecting the midpoints of ABCD. Z(a, b) ) b ⫺b N T TP , MN 6 equal ) ( ( ( 2b )2 2 Because the diagonals are congruent, parallelogram XYZW is a A M ( ( a a XZ ⫽ YW opposite sides of the trapezoid. MN 6 , then the congruent from Theorem 6–14. rectangle Usethe diagram from the Example on page 82. Explain why the proof using A(2a, 0), B(0, 2b), C(2a, 0), and D(0, 2b) is easier than the proof using A(a, 0), B(0, b), C(a, 0), and D(0, b). Using multiples of 2 in the coordinates for A, B, C, and D eliminates the use of fractions when finding midpoints, since findingmidpoints requires division by 2. X(a, b) Y(a, b) B(0, 2b) From Lesson 6-6, you know that XYZW is a parallelogram. 82 Geometry Lesson 6-7 Daily Notetaking Guide L1 Name_____________________________________ Class____________________________ Date________________ © PearsonEducation, Inc., publishing as Pearson Prentice Hall. 83 Name_____________________________________ Class____________________________ Date ________________ Ratios and Proportions Lesson Objective 1 Write ratios and solve proportions a. NAEP 2005 Strand: Geometry Topic: Positionand Direction Properties of Proportions b d b (4) a 1 b c1d d All rights reserved. (3) ac All rights reserved. d c A proportion is a statement that two ratios are equal. ) Cross-Product Property Simplify. n5 14 Divide each side by 5 . ( 2x x 1 1 1 2 x5 2 ) Cross-Product Property Distributive Property Subtract 2xfrom each side. Quick Check. 1. A photo that is 8 in. wide and 531 in. high is enlarged to a poster that is 2 ft c a b d and a : b c : d are examples of proportions. extended proportion 35 70 3x 5 c a b d is equivalent to (2) ba ( 5n 5 1 x b. x 1 3 52 3x 5 2 Vocabulary and Key Concepts. bc 25 n 5 35 5n 5 2Local Standards: ____________________________________ An Geometry Lesson 6-7 2 Solving for a Variable Solve each proportion. Lesson 7-1 (1) ad Daily Notetaking Guide L1 wide and 113 ft high. What is the ratio of the height of the photo to the height is a statement that three or more ratios areequal. of the poster? 6 1 4 24 16 4 is an example of an extended proportion. The Cross-Product Property states that the product of the extremes of a proportion is equal 1:3 a c b d a d b c extremes A scale drawing is a drawing in which all lengths are proportional to corresponding actual lengths. A scale isthe ratio of any length in a scale drawing to the corresponding actual length. The lengths may be in different units. Examples. 1 Finding Ratios A scale model of a car is 4 in. long. The actual car is 15 ft long. What is the ratio of the length of the model to the length of the car? Write both measurements in thesame units. 180 15 ft 15 12 in. in. length of model 4 in. 4 in. 4 length of car 5 15 ft 5 180 in. 5 180 5 L1 Geometry Lesson 7-1 2. Solve each proportion. a. 5z 5 20 3 z 0.75 6 b. n 18 165n n3 1 45 The ratio of the length of the scale model to the length of the car is 84 © Pearson Education, Inc., publishing asPearson Prentice Hall. means a:b c:d © Pearson Education, Inc., publishing as Pearson Prentice Hall. to the product of the means. 1 : 45 . Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 7-1 Geometry 85 21 Geometry: All-In-One Answers Version B(continued) Name_____________________________________ Class____________________________ Date ________________ Lesson 7-2 2 Using Similar Figures If 䉭ABC 䉭YXZ, find the value of x. NAEP 2005 Strand: Geometry and Measurement Identify similar polygons Apply similar polygonsBecause 䉭ABC 䉭YXZ, you can write and solve a proportion. Topics: Transformation of Shapes and Preservation of Properties; Measuring Physical Attributes Local Standards: ____________________________________ AC 5 YZ BC Corresponding sides are XZ proportional 40 Vocabulary. x Similarfigures have the same shape but not necessarily the same size. Two polygons are All rights reserved. is ⬃. similarity The mathematical symbol for The similarity ratio is the ratio of the lengths of corresponding sides of similar figures. A All rights reserved. 50 similar if corresponding angles are congruentand corresponding sides are proportional. is a rectangle that can be divided into a square and a golden rectangle x The golden ratio is the ratio of the length to the width of any golden rectangle, about Z 40 12 Substitute. 50 Solve for x. 15 Simplify. B Complete each statement. a. m⬔B Y A ⬔B ⬔ Y andm⬔Y 78, so m⬔B 78 because congruent angles have the same measure. C X BC 5 b. YZ XZ 78 42 Z BC AC . corresponds to XZ, so YZ XZ Quick Check. 1. Refer to the diagram for Example 1. Complete: 42 BC and YZ All rights reserved. Example. © Pearson Education, Inc., publishing as PearsonPrentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 9 1 Understanding Similarity 䉭ABC 䉭XYZ. 86 12 2. Refer to the diagram for Example 2. Find AB. 1.618 : 1. m⬔A . 50 X C Quick Check. rectangle that is similar to the original rectangle. AC B 30 x 12 40 x5 Y A AB XYGeometry Lesson 7-2 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date ________________ Daily Notetaking Guide Name_____________________________________ Class____________________________ Date________________ W 2 Using Similarity Theorems Explain why the triangles Lesson 7-3 Proving Triangles Similar Lesson Objectives 1 Use AA, SAS, and SSS similarity statements 2 Apply AA, SAS, and SSS similarity statements 87 Geometry Lesson 7-2 must be similar. Write a similarity statement. 24⬔YVZ > ⬔WVX because they are vertical angles. NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties VY 12 VW 5 24 1 2 VZ 5 and VX 18 36 Z 18 1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 Example. Similar Polygons LessonObjectives 2 Name_____________________________________ Class____________________________ Date________________ V , 2 12 Local Standards: ____________________________________ 36 Y so corresponding sides are proportional. X Vocabulary and Key Concepts. 䉭TRS 䉭 R PLM PT S L M All rights reserved. Postulate 7-1: Angle-Angle Similarity (AA⬃) Postulate If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. All rights reserved. Therefore, 䉭YVZ 䉭WVX Side-Angle-Side Similarity (SAS) by the Theorem. Quick Check. 1. InExample 1, you have enough information to write a similarity statement. Do you have enough information to find the similarity ratio? Explain. No; none of the side lengths are given. Theorem 7-1: Side-Angle-Side Similarity (SAS⬃) Theorem If an angle of one triangle is congruent to an angle of a secondtriangle, and the sides including the two angles are proportional, then © Pearson Education, Inc., publishing as Pearson Prentice Hall. the triangles are similar. Indirect measurement is a way of measuring things that are difficult to measure directly. Examples. 1 Using the AA⬃ Postulate MX ' AB. Explainwhy the triangles M are similar. Write a similarity statement. Because MX ' AB, ⬔AXM and ⬔BXK are right angles , so ⬔AXM > ⬔ BXK K . ⬔A > ⬔ B because their measures are equal. 䉭AMX 䉭BKX Angle-Angle Similarity (AA) by the 88 22 Geometry Lesson 7-3 Geometry Postulate. © PearsonEducation, Inc., publishing as Pearson Prentice Hall. the triangles are similar. Theorem 7-2: Side-Side-Side Similarity (SSS⬃) Theorem If the corresponding sides of two triangles are proportional, then 2. Explain why the triangles at the right must be similar. Write a similarity statement. G 8 AC 5 CB 5 AB 53 , so the triangles are similar by the SSS~ Theorem; EF 4 EG GF nABC , nEFG 8 12 E F 9 A 6 B 6 C 58 A 58 X Daily Notetaking Guide B L1 L1 Daily Notetaking Guide Geometry Lesson 7-3 All-In-One Answers Version B 89 L1 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date ________________ Name_____________________________________ Class____________________________ Date________________ Lesson 7-4 2 Finding Distance At a golf course, Maria drove herball 192 yd straight Similarity in Right Triangles Lesson Objective 1 Find and use relationships in similar right triangles Topic: Transformation of Shapes and Preservation of Properties x 192 240 Local Standards: ____________________________________ 192x The altitude to the hypotenuse of a righttriangle divides the triangle into Write a proportion. similar to the original triangle and to each other. 240 x M 192 Cross-Product Property 57,600 36,864 G 192 192 x All rights reserved. All rights reserved. Theorem 7-3 similar 240 192 (x 1 192) 5 2402 Vocabulary and Key Concepts. two triangles that areCup y toward the cup. Her brother Gabriel drove his ball straight 240 yd, but not toward the cup. The diagram shows the results. Find x and y, their remaining distances from the cup. Use Corollary 2 of Theorem 7-3 to solve for x. NAEP 2005 Strand: Geometry ★ Distributive Property 20,736 Solve for x. x 5108 Use Corollary 2 of Theorem 7-3 to solve for y. x 192 Corollary 1 to Theorem 7-3 y x y Write a proportion. The length of the altitude to the hypotenuse of a right triangle is the geometric mean 108 192 of the lengths of the segments adjacent hypotenuse segment All rights reserved. and the length of the. hypotenuse The geometric mean of two positive numbers a and b is the positive number x such a 5 x. that x b Examples. 1 Finding the Geometric Mean Find the geometric mean of 3 and 12. x 3 x5 Write a proportion. 12 x2 5 36 x5 x5 6 © Pearson Education, Inc., publishing as Pearson Prentice Hall. ©Pearson Education, Inc., publishing as Pearson Prentice Hall. so that the length of each leg of the triangle is the geometric mean of the length of the y y The altitude to the hypotenuse of a right triangle separates the hypotenuse Substitute 108 for x. 108 300 Corollary 2 to Theorem 7-3 y y of thehypotenuse. Simplify. 108 y2 32,400 Cross-Product Property !32,400 y Find the positive square root. y 180 Maria’s ball is yd from the cup, and Gabriel’s ball is 108 yd 180 from the cup. Quick Check. 1. Find the geometric mean of 5 and 20. 2. Recall Example 2. Find the distance between Maria’s ball andGabriel’s ball. 144 yd 10 Cross-Product Property. !36 Find the positive square root. The geometric mean of 3 and 12 is 6 . 90 Geometry Lesson 7-4 Daily Notetaking Guide L1 Name_____________________________________ Class____________________________ Date ________________Geometry Lesson 7-4 91 Name_____________________________________ Class____________________________ Date________________ 2 Using the Triangle-Angle-Bisector Theorem Find the value of x. Lesson 7-5 Proportions in Triangles Lesson Objectives 1 Use the Side-Splitter Theorem 2Use the Triangle-Angle-Bisector Theorem © Pearson Education, Inc., publishing as Pearson Prentice Hall. Daily Notetaking Guide L1 G 40 24 NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties I x Local Standards: ____________________________________IG GH Key Concepts. 24 then it divides those sides proportionally. All rights reserved. All rights reserved. If a line is parallel to one side of a triangle and intersects the other two sides, Theorem 7-5: Triangle-Angle-Bisector Theorem 24 ( 40 IK Triangle-Angle-Bisector Theorem. HK x Substitute. 30 40Theorem 7-4: Side-Splitter Theorem H 30 K 30 ) x x Solve for x. 18 Quick Check. If a ray bisects an angle of a triangle, then it divides the opposite side into two segments that 1. Use the Side-Splitter Theorem to find the value of x. are proportional to the other two sides of the triangle. 1.5 5 x 1.5 2.5 xtransversals are proportional. a c b d a b5 c d Examples. 1 Using the Side-Splitter Theorem Find y. B CM MB 12 y 10 92 L1 y 12 M CN 10 Side-Spitter Theorem N NA A 10 C 6 © Pearson Education, Inc., publishing as Pearson Prentice Hall. If three parallel lines intersect two transversals, then thesegments intercepted on the © Pearson Education, Inc., publishing as Pearson Prentice Hall. Corollary to Theorem 7-4 2. Find the value of y. 3.6 5.76 y 5 8 Substitute. 6 y 72 Cross-Product Property y 7.2 Solve for y. Geometry Lesson 7-5 Daily Notetaking Guide All-In-One Answers Version B L1 L1 DailyNotetaking Guide Geometry Lesson 7-5 Geometry 93 23 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ 2 Is It a Right Triangle? Is this triangle a right triangle? Lesson 8-1 1 4mThe Pythagorean Theorem and Its Converse a2 1 b2 0 c2 Lesson Objectives NAEP 2005 Strand: Geometry Use the Pythagorean Theorem Use the Converse of the Pythagorean Theorem 2 Name_____________________________________ Class____________________________ Date________________ Topic: Relationships Among Geometric Figures 42 62 0 72 Substitute 4 for a, 6 for b, and 7 for c. 16 36 0 49 Simplify. 52 2 49 Local Standards: ____________________________________ Because a2 b2 ⫽ c2, the triangle Vocabulary and Key Concepts. Theorem 8-5: PythagoreanTheorem 7m 6m a right triangle. is not c a . b a2 b2 c 2 Theorem 8-6: Converse of the Pythagorean Theorem All rights reserved. hypotenuse is equal to the square of the length of the All rights reserved. Quick Check. legs In a right triangle, the sum of the squares of the lengths of the 1. A right triangle hasa hypotenuse of length 25 and a leg of length 10. Find the length of the other leg. Do the lengths of the sides form a Pythagorean triple? 5!21; no If the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle. 2.A triangle has sides of lengths 16, 48, and 50. Is the triangle a right triangle? A Pythagorean triple is a set of nonzero whole numbers a, b, and c that satisfy B Do the lengths of the sides of #ABC form a Pythagorean triple? 1 a2 b2 5 c2 Use the 52 122 c2 Î 169 Pythagorean Theorem. 12 Substitute 5 for a,and 12 for b. 169 c2 Simplify. c Take the square root. 13 c A C 5 Simplify. The length of the hypotenuse is 13 . The lengths of the sides, 5, 12, and 13 form a Pythagorean triple because they are whole All rights reserved. Examples. 1 Pythagorean Triples Find the length of the hypotenuse of #ABC. no ©Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. the equation a 2 ⫹ b 2 ⫽ c 2. numbers that satisfy a2 1 b2 5 c2. Geometry Lesson 8-1 Daily Notetaking Guide L1 L1 Name_____________________________________Class____________________________ Date________________ Lesson 8-2 Geometry Lesson 8-1 2 Using the Length of One Side Find the value of each variable. Use the 30-60-90 Triangle Theorem to find the lengths of the legs. 4 !3 2 ? x NAEP 2005 Strand: Geometry Topic: Relationships AmongGeometric Figures Local Standards: ____________________________________ x Key Concepts. hypotenuse ⴝ 2 ? shorter leg 4"3 Divide each side by 2 . "3 y "3 ? 2 !3 s2 hypotenuse "2 ? 45 s 45 s leg y 2 ? "3 ? "3 All rights reserved. and the All rights reserved. congruent length of the hypotenuse is "2times the length of a leg. y y Simplify. 60 y "3 ? shorter leg Theorem 8-5: 45-45-90 Triangle Theorem 30 4 3 2 x 2 In a 45-45-90 triangle, both legs are 95 Name_____________________________________ Class____________________________ Date ________________ Special Right TrianglesLesson Objectives 1 Use the properties of 45-45-90 triangles 2 Use the properties of 30-60-90 triangles Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. 94 30°-60°-90° Substitute 2"3 Triangle Theorem x for shorter leg. Simplify. 6 The length of the shorter leg is 2"3 ,and the length of the longer leg is 6 . Quick Check. 1. Find the length of the hypotenuse of a 45-45-90 triangle with legs of length 5 !3. Theorem 8-6: 30-60-90 Triangle Theorem twice the length of the hypotenuse 2 ? longer leg "3 ? 5!6 . The length of the longer leg is "3 times the length of the shorter leg 2s. shorter leg 30 s3 60 s shorter leg © Pearson Education, Inc., publishing as Pearson Prentice Hall. shorter leg Examples. 1 Finding the Length of the Hypotenuse Find the value of the variable. Use the 45-45-90 Triangle Theorem to find the hypotenuse. h "2 ? 5!6 "12 h 5 h5 hypotenuse ⴝ "2 45 h ? leg 5 6Simplify. 45 4(3) ( ) 3 5 6 h5 2 h 10 "3 The length of the hypotenuse is 96 24 Geometry Lesson 8-2 Geometry 10"3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. In a 30-60-90 triangle, the length of the hypotenuse is 2. Find the lengths of the legs of a 30-60-90 triangle with hypotenuse oflength 12. 6; 6!3 . Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 8-2 All-In-One Answers Version B 97 L1 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________2 Using a Tangent Ratio To measure the height of a tree, Alma walked Lesson 8-3 The Tangent Ratio Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ 125 ft from the tree and measured a 32 angle from theground to the top of the tree. Estimate the height of the tree. NAEP 2005 Strand: Measurement Use tangent ratios to determine side lengths in triangles Topic: Measuring Physical Attributes Local Standards: ____________________________________ 32 125 ft Vocabulary. A Leg ⬔A C adjacent to Allrights reserved. All rights reserved. B opposite 125 ⬔A . All rights reserved. B 29 98 A opposite adjacent opposite adjacent BC AC AC BC 20 21 21 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Examples. tan B Solve forheight. 78.108669 32 78 Use a calculator. feet tall. J 3, 7 7 3 3 L K 7 2. Find the value of w to the nearest tenth. a. b. 10 54 w 1.0 28 w 13.8 1.9 20 Geometry Lesson 8-3 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________Date________________ Daily Notetaking Guide 99 Geometry Lesson 8-3 Name_____________________________________ Class____________________________ Date ________________ 2 Using the Cosine Ratio A 20-ft wire supporting a flagpole forms a Lesson 8-4 Sine and Cosine RatiosLesson Objective 1 Use sine and cosine to determine side lengths in triangles © Pearson Education, Inc., publishing as Pearson Prentice Hall. ) 1. Write the tangent ratios for ⬔K and ⬔J. 1 Writing Tangent Ratios Write the tangent ratios for ⬔A and ⬔B. tan A tan 32° Quick Check. opposite adjacent Youcan abbreviate the equation as tan A 21 Use the tangent ratio. 125 ( height 125 length of leg opposite lA length of leg adjacent to lA tangent of ⬔A C height tan 32 The tree is about 20 angle with the ground, so you can use the tangent ratio to estimate the height of the tree. lA to the length of the legadjacent to lA. Leg right The tree forms a The tangent of acute ⬔A in a right triangle is the ratio of the length of the leg opposite 35 angle with the flagpole. To the nearest foot, how high is the flagpole? NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes The flagpole, wire, and groundform a Local Standards: ____________________________________ measure of its hypotenuse, you can use the wire as the hypotenuse right triangle 35 20 ft with the . Because you know an angle and the cosine ratio to find the height of the flagpole. Vocabulary. This can be abbreviated sin A cosine oflA B leg opposite lA hypotenuse hypotenuse opposite hypotenuse opposite . leg adjacent to /A hypotenuse This can be abbreviated cos A Leg cos T sin G hypotenuse adjacent hypotenuse opposite hypotenuse adjacent cos G hypotenuse 100 L1 Geometry Lesson 8-4 20 16 T 12 20 16 20 4 R 12 G 4 520 16 Solve for height. Use a calculator. feet tall. X 48 80 Z 5 20 16 16.383041 ⬔A 3 5 35 cos 35° 1. Write the sine and cosine ratios for ⬔X and ⬔Y. sin G, and cos G. Write your answers in simplest terms. sin T ? 48 48 64 sin X ⫽ 64 80, cos X ⫽ 80, sin Y ⫽ 80, cos Y ⫽ 80 1 Writing Sine and CosineRatios Use the triangle to find sin T, cos T, 12 20 The flagpole is about adjacent to adjacent hypotenuse. Examples. opposite 20 Use the cosine ratio. 20 height Quick Check. ⬔A C A Leg 3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. sine of ⬔A All rights reserved. of ⬔A is the ratio ofthe length of the leg adjacent to ⬔A to the hypotenuse. © Pearson Education, Inc., publishing as Pearson Prentice Hall. cosine All rights reserved. hypotenuse. The height cos 35 The sine of ⬔A is the ratio of the length of the leg opposite ⬔A to the length of the 64 Y 2. In Example 2, suppose that theangle the wire makes with the ground is 50°. What is the height of the flagpole to the nearest foot? 15 feet 20 ft 50 5 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 8-4 Geometry 101 25 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ Lesson 8-5 2 Aviation An airplane flying 3500 ft above the ground begins a 2 descent to land at the airport. How many miles from the airport is the airplane when it starts its descent?(Note: The angle is not drawn to scale.) NAEP 2005 Strand: Measurement Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ Angles of Elevation and Depression Topic: Measuring Physical Attributes Use angles ofelevation and depression to solve problems 2 x Local Standards: ____________________________________ 3,500 ft Ground ` Vocabulary. An angle of elevation is the angle formed by a horizontal line and the line of sight to an sin 2⬚ 3,500 x is the angle formed by a horizontal line and the line of angleof depression sight to an object below the horizontal line. Horizontal line Angle of B All rights reserved. An All rights reserved. object above the horizontal line. 5280 100287.9792 Angle of depression 2 3500 Use the ratio. sine 3500 x Solve for x. sin 2⬚ 100287 .9792 18.993935 Use a calculator. 5280Divide by to convert feet to miles. elevation A The airplane is about Horizontal line 19 miles from the airport when it starts its descent. Quick Check. 1 2 3 4 One side of the angle of depression is a horizontal line. ⬔1 is the angle of depression airplane from the building to the . 1. Describe each angle as itrelates to the situation in Example 1. a. ⬔3 The angle of depression from the building to the person on the ground All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. as they relate to the situation shown. © Pearson Education, Inc., publishing as Pearson Prentice Hall.Examples. 1 Identifying Angles of Elevation and Depression Describe ⬔1 and ⬔2 b. ⬔4 The angle of elevation from the person on the ground to the building 2. An airplane pilot sees a life raft at a 26 angle of depression. The airplane’s altitude is 3 km. What is the airplane’s surface distance d from theraft? about 6.8 km One side of the angle of elevation is a horizontal line. ⬔2 is the angle of 102 from the to the building . airplane Geometry Lesson 8-5 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ 2 Adding Vectors Vectors v 4, 3 and w 4, 3 are shown at the right. Lesson 8-6 Vectors Lesson Objectives 1 Describe vectors 2 Solve problems that involvevector addition NAEP 2005 Strand: Geometry Topic: Position and Direction To find the first coordinate of s, add the first y 4 Write the sum of the two vectors as an ordered pair. Then draw s, the sum of v and w. 2 (4, 3) coordinates of 2 v and w. Local Standards:____________________________________ Vocabulary and Key Concepts. To find the second coordinate of s, add the second coordinates of 2 v and w. 4 8x 4 6 (4, 3) s k4, 3l k4, 3l Adding Vectors k All rights reserved. . A vector is any quantity with magnitude (size) and direction. 4ⴙ4 , 3 ⴙ (ⴚ3) k 8 , 0 lAll rights reserved. kx1 1 x2, y1 1 y2 l For a x1, y1 and c x2, y2, a c 4 2 Draw vector v using the origin as the initial point. Draw vector w using the terminal point of v, (4, 3), as the initial point. Draw the can be represented with an arrow. vector l Add the coordinates. Simplify. resultant vector s from the A103 Geometry Lesson 8-5 © Pearson Education, Inc., publishing as Pearson Prentice Hall. elevation the terminal point initial point of w. of v to y v 2 w 4 s8 x 6 2 4 The magnitude of a vector is its size, or length. Quick Check. initial point The of a vector is the point at which it starts. is the sum of othervectors. y Examples. 1 Describing a Vector Describe OM as an ordered pair. x Give coordinates to the nearest tenth. Use the sine and cosine ratios to find the values of x and y. cos 40 x Use sine and cosine. 80 x 80 ( cos 40ⴗ x ) 61.28355545 Because point M is in the negative 104 26 sin 40 Solve for thevariable. Use a calculator. third Geometry y y 80 M 80 y 80( sin 40ⴗ y x ) 51.42300878 51 k221.6, 46.2l 65 O x 2. Write the sum of the two vectors k2, 3l and k24, 22l as an ordered pair. k22, 1l quadrant, both coordinates are . To the nearest tenth, OM Geometry Lesson 8-6 40 O © Pearson Education,Inc., publishing as Pearson Prentice Hall. resultant vector © Pearson Education, Inc., publishing as Pearson Prentice Hall. A y 1. Describe the vector at the right as an ordered pair. Give the coordinates to the nearest tenth. The terminal point of a vector is the point at which it ends. k261.3, 251.4l . DailyNotetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 8-6 All-In-One Answers Version B 105 L1 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ 2 Finding aTranslation Image Find the image of 䉭ABC Lesson 9-1 1 A(1, 2) → A( ⴚ1 ⴙ 2 , 2 3) NAEP 2005 Strand: Geometry Identify isometries Find translation images of figures y under the translation (x, y) → (x 2, y 3). Translations Lesson Objectives 2 Name_____________________________________Class____________________________ Date ________________ A Use the rule (x, y ) → (x ⴙ 2, y ⴚ 3) Topic: Transformation of Shapes and Preservation of Properties; Position and Direction B(1, 0) → B( 1 2, 0 ⴚ 3 ) Local Standards: ____________________________________ The image of 䉭ABC is䉭ABC with A( 1 , ⴚ1 ), B( 3 , ⴚ3 ), x B C C(0, 1) → C( 0 ⴙ 2 , ⴚ1 ⴚ 3 ) and C( 2 , ⴚ4 ). Vocabulary. A transformation of a geometric figure is a change in its position, shape, or size. Quick Check. is the original image before changes are made. In a transformation, the image is the resulting figure afterchanges are made. An isometry All rights reserved. preimage All rights reserved. In a transformation, the 1. Does the transformation appear to be an isometry? Explain. a. b. Preimage Preimage Image Yes; the figures appear to be congruent by a flip and a slide. Yes; the figures appear to be congruent bya flip. is a transformation in which the preimage and the image Image are congruent. A translation (slide) is a transformation that maps all points the same distance and in the 2. Find the image of 䉭ABC for the translation (x 2, y 1). same direction. Aⴕ(ⴚ3, 3), Bⴕ(ⴚ1, 1), Cⴕ(ⴚ2, 0) is a combination of two ormore transformations. All rights reserved. Examples. 1 Identifying Isometries Does the transformation appear to be an isometry? Image Preimage The image appears to be the same as the preimage, but Because the figures appear to be congruent turned . , the transformation appears to be an isometry.106 Geometry Lesson 9-1 L1 L1 Daily Notetaking Guide 107 Geometry Lesson 9-1 Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date________________ Lesson 9-2 Examples. Reflections Lesson Objectives 1 Find reflection images of figures © Pearson Education, Inc., publishing as Pearson Prentice Hall. Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. composition of transformations © PearsonEducation, Inc., publishing as Pearson Prentice Hall. A 2 Drawing Reflection Images 䉭XYZ has vertices X(0, 3), Y(2, 0), NAEP 2005 Strand: Geometry Topics: Transformation of Shapes and Preservation of Properties y and Z(4, 2). Draw 䉭XYZ and its reflection image in the line x 4. First locate verticesX, Y, and Z and draw 䉭XYZ in a coordinate plane. Local Standards: ____________________________________ Locate points X, Y, and Z such that the line of reflection x 4 perpendicular bisector is the Vocabulary. x4 ‚ 4 X 2 O 2 of XXr, YYr, and ZZr. X Z Z Y 6 Y ‚ ‚ 8x 2 Draw the reflection image XYZ.itself B is the point such that r is the , and if a point B is not on line r, then its image perpendicular bisector B Line of All rights reserved. the image of A is of BBr. r A Aⴕ All rights reserved. A reflection in line r is a transformation such that if a point A is on line r, then 3 Congruent Angles D is the reflection ofD across ᐉ. Show that PD and PW form congruent angles with line ᐉ. isometry Because a reflection is an D W , PD and PDr form congruent angles with line ᐉ. P PDr and PW also form congruent angles with line ᐉ reflection because Bⴕ 艎 angles are congruent. vertical Therefore, PD and PW formcongruent angles with line ᐉ by the Property. Transitive D Example. 1 Finding Reflection Images If point Q(1, 2) is reflected across y x=1 Quick Check. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Q Q is 2 units to the left of the reflection line, so its image Q is x ⴚ2 2 units to the right ofthe reflection line. The reflection line is the 2 perpendicular bisector of QQr if Q is at (3, 2) . Quick Check. 1. What are the coordinates of the image of Q if the reflection line is y 1? (ⴚ1, ⴚ4) 108 L1 Geometry Lesson 9-2 Daily Notetaking Guide All-In-One Answers Version B L1 © Pearson Education, Inc.,publishing as Pearson Prentice Hall. line x 1, what are the coordinates of its reflection image? 2. 䉭ABC has vertices A(3, 4), B(0, 1), and C(2, 3). Draw 䉭ABC and its reflection image in the line x 3. y 8 A Aⴕ 4 C Cⴕ Bⴕ B ⴚ8 ⴚ4 4 8 12 x ⴚ4 ⴚ8 3. In Example 3, what kind of triangle is 䉭DPD? Imagine theimage of point W reflected across line ᐉ. What can you say about 䉭WPW and 䉭DPD? isosceles; they are similar by SAS L1 Daily Notetaking Guide Geometry Lesson 9-2 Geometry 109 27 Geometry: All-In-One Answers Version B (continued) Name_____________________________________Class____________________________ Date________________ A 2 Identifying a Rotation Image Regular hexagon ABCDEF is divided Lesson 9-3 Rotations Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ intosix equilateral triangles. a. Name the image of B for a 240 rotation about M. NAEP 2005 Strand: Geometry Draw and identify rotation images of figures Because 360 6 60 , each central angle of ABCDEF measures 60° . A 240 counterclockwise rotation about center M moves point B across triangles. Theimage of point B is point D . four Topic: Transformation of Shapes and Preservation of Properties Local Standards: ____________________________________ B F C M D E b. Name the image of M for a 60° rotation about F. Vocabulary. 䉭AMF is equilateral, so ⬔AFM has measure 180 3 A rotation of xabout a point R is a transformation for which the following are true: RV R R ). R and m⬔VRV x x . V All rights reserved. • For any point V, RV All rights reserved. (that is, R itself • The image of R is 60 . A 60 60 rotation of 䉭AMF about point F would superimpose FM on FA, so the V image of M under a 60rotation about point F is point A . Quick Check. 1. Draw the image of LOB for a 50 rotation about point B. Label the vertices of the image. Examples. 1 Drawing a Rotation Image Draw the image of 䉭LOB under a 60 rotation about C. Step 1 Use a protractor to draw a 60 angle at vertex C with one side CO.O Step 2 Use a compass to construct COr CO. L Oⴕ L L L B B C O O L B Lⴕ C O Bⴕ O L B 110 B C O B O L B L Geometry Lesson 9-3 L L1 2. Regular pentagon PENTA is divided into 5 congruent triangles. Name the image of T for a 144 rotation about point X. E L1Name_____________________________________ Class____________________________ Date________________ P A Daily Notetaking Guide E X Geometry Lesson 9-3 N 111 Name_____________________________________ Class____________________________ Date ________________Symmetry Lesson Objective 1 Identify the type of symmetry in a figure L T Daily Notetaking Guide Lesson 9-4 B B O All rights reserved. C O 60° Examples. 1 Identifying Lines of Symmetry Draw all lines of symmetry for the NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservationof Properties © Pearson Education, Inc., publishing as Pearson Prentice Hall. C O © Pearson Education, Inc., publishing as Pearson Prentice Hall. C O © Pearson Education, Inc., publishing as Pearson Prentice Hall. Step 3 Locate L⬘ and B⬘ in a similar manner. Then draw 䉭L⬘O⬘B⬘. isoscelestrapezoid. Local Standards: ____________________________________ Draw any lines that divide the isosceles trapezoid so that half of the figure is a mirror image of the other half. Vocabulary. reflectional symmetry if there is symmetry that maps the figure onto itself. Line symmetry is the same asreflectional symmetry. All rights reserved. A figure has All rights reserved. A figure has symmetry if there is an isometry that maps the figure onto itself. There is one line of symmetry. 2 Identifying Rotational Symmetry Judging from appearance, do the letters V and H have rotational symmetry? If so, givean angle of rotation. The letter V does not have rotational symmetry because it must be rotated 360 Line of Symmetry before it is its own image. The letter H is its own image after one half-turn, so it has rotational A figure has © Pearson Education, Inc., publishing as Pearson Prentice Hall. ( reflectional )rotational symmetry if it is its own image for some rotation of 180° or less. A figure has point symmetry if it has 180⬚ rotational symmetry. The figure is its own image after one half-turn, so it has rotational symmetry with a The figure also has point 180 angle of rotation. symmetry. © Pearson Education,Inc., publishing as Pearson Prentice Hall. symmetry with a The heart-shaped figure has symmetry. line 180 angle of rotation. Quick Check. 1. Draw a rectangle and all of its lines of symmetry. 2. a. Judging from appearance, tell whether the figure at the right has rotational symmetry. If so, give the angle ofrotation. yes; 180° b. Does the figure have point symmetry? yes 112 28 Geometry Lesson 9-4 Geometry Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 9-4 All-In-One Answers Version B 113 L1 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ 2 Scale Drawings The scale factor on a museum’s floor plan is 1 : 200. The Lesson 9-5 Dilations Lesson Objective 1 Name_____________________________________Class____________________________ Date ________________ length and width of one wing on the drawing are 8 in. and 6 in. Find the actual dimensions of the wing in feet and inches. NAEP 2005 Strand: Geometry Locate dilation images of figures 1 Topic: Geometry The floor plan is a reduction ofthe actual dimensions by a scale factor of 200 . Local Standards: ____________________________________ Multiply each dimension on the drawing by 200 to find the actual dimensions. Then write the dimensions in feet and inches. Vocabulary. A dilation is a transformation with center C and scalefactor n for which the following are true: R 1 C CR . R 3 Q All rights reserved. ) (that is, C C ). • For any point R, R is on CR and CR n ft y J J dilation of concentric circle B with 8-cm diameter. Describe the dilation. The circles are concentric, so the dilation has center C . Because the diameter of the dilationimage is smaller, the dilation is a . diameter of dilation image diameter of preimage 5 3 8 3 8 The dilation is a reduction with center C and scale factor . Geometry Lesson 9-5 Daily Notetaking Guide © Pearson Education, Inc., publishing as Pearson Prentice Hall. Examples. 1 Finding a Scale Factor CircleA with 3-cm diameter and center C is a reduction Lesson 9-6 by in. 100 ft . 2 K x K 4 L L The dilation is a reduction with center (0, 0) and scale factor 1 2. 2. The height of a tractor-trailer truck is 4.2 m. The scale factor for a model 1 . Find the height of the model to the nearest centimeter. truck is 54 8 cmL1 L1 Name_____________________________________ Class____________________________ Date________________ Daily Notetaking Guide Geometry Lesson 9-5 115 Name_____________________________________ Class____________________________ Date ________________Example. Compositions of Reflections Lesson Objectives 1 Use a composition of reflections 2 Identify glide reflections O M M 3 is a dilation with a scale factor less than 1. reduction 133 ft, 4 in. 4 1. Quadrilateral JKLM is a dilation image of quadrilateral JKLM. Describe the dilation. 2 © Pearson Education,Inc., publishing as Pearson Prentice Hall. All rights reserved. ft, 100 with center C and scale factor 3 . A © Pearson Education, Inc., publishing as Pearson Prentice Hall. 133 in. dilation RrQr is the image of RQ under a An enlargement is a dilation with a scale factor greater than 1. 114 in. 1200 QuickCheck. Q Scale factor: 1600 6 in. 200 The museum wing measures All rights reserved. itself • The image of C is 8 in. 200 Composition of Reflections in Intersecting Lines The letter D is reflected in line x and then in line y. Describe the resulting rotation. Find the image of D through a reflection across linex. Find the image of the reflection through another reflection across line y. NAEP 2005 Strand: Geometry Topic: Transformation of Shapes and Preservation of Properties Local Standards: ____________________________________ D D Vocabulary and Key Concepts. D 43 A y reflections . Theorem 9-2

A composition of reflections across two parallel lines is a . translation All rights reserved. x The composition of two reflections across intersecting lines is a The center of rotation is is twice the angle formed by the intersecting lines 86 clockwise, or Theorem 9-3 is rotated A composition of reflections acrosstwo intersecting lines is a rotation at point A . rotation . R R R R R A glide reflection is the composition of a glide (translation) and a reflection across a line © Pearson Education, Inc., publishing as Pearson Prentice Hall. R R R R Glide reflection Rotation L1 Geometry Lesson 9-6 c b. Use parallel lines ᐉand m. Draw R between ᐉ and m. Find the image of R for a reflection across line ᐉ and then across line M. Describe the resulting translation. All-In-One Answers Version B L1 R Answers may vary. The result is a translation twice the distance between ᐉ and m. R 艎 Daily Notetaking Guide b R parallel tothe direction of translation. 116 a Answers may vary. The result is a clockwise rotation about point c through an angle of 2mlacb. R © Pearson Education, Inc., publishing as Pearson Prentice Hall. Translation . So, the letter D counterclockwise, with the center of R . Theorem 9-5: Isometry ClassificationTheorem There are only four isometries. They are the following: Reflection 274 a. Reflect the letter R across a and then b. Describe the resulting rotation. In a plane, one of two congruent figures can be mapped onto the other by a reflections . Quick Check. Theorem 9-4: Fundamental Theorem ofIsometries composition of at most three rotation the point where the lines intersect , and the angle R A translation or rotation is a composition of two All rights reserved. Theorem 9-1 L1 Daily Notetaking Guide m Geometry Lesson 9-6 Geometry 117 29 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ 2 Identifying Symmetries in Tessellations List the Lesson 9-7 Tessellations Lesson Objectives 1 2 Name_____________________________________Class____________________________ Date ________________ Identify transformation in tessellations, and figures that will tessellate Identify symmetries in tessellations symmetries in the tessellation. NAEP 2005 Strand: Geometry Starting at any vertex, the tessellation can be mapped Topic:Geometry onto itself using a Local Standards: ____________________________________ has point 180⬚ rotation, so the tessellation symmetry centered at any vertex. translational The tessellation also has Vocabulary and Key Concepts. symmetry, as can be seen by sliding any triangle Theorem 9-6Every triangle tessellates. All rights reserved. All rights reserved. © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. onto a copy of itself along any of the lines. Theorem 9-7 Every quadrilateral tessellates. A tessellation, or tiling,is a repeating pattern of figures that completely covers a plane, Quick Check. 1. Identify a transformation and outline the smallest repeating figure in the tessellation below. without gaps or overlaps. is the type of symmetry for which there is a translation that Translational symmetry maps a figure onto itself.maps a figure onto itself. Examples. 1 Identifying the Transformation in a Tessellation Identify the repeating figures and a transformation in the tessellation. octagon and one adjoining will completely cover the plane without gaps or square overlaps. Use arrows to show a translation. 118 Geometry Lesson9-7 Daily Notetaking Guide L1 The height corresponding to the 9-in. base is 15 in. Find the area of the parallelogram. A bh Local Standards: ____________________________________ h . b bh Theorem 10-2: Area of a Parallelogram height and the corresponding A h . b of a parallelogram correspondingto a given base is the segment perpendicular to the line Altitude Base containing that base drawn from the side opposite the base. The height of a parallelogram is the length of its altitude. A base of a triangle is any of its sides. © Pearson Education, Inc., publishing as Pearson Prentice Hall. height 1bh 2altitude for h. 135 in.2 A 5 12bh A 5 12 ( Area of a 30 )( 13 ) 195 Substitute X triangle 30 for b and 13 31 cm for h. 13 cm Simplify. 195 cm2. Y Z 30 cm 1. A parallelogram has sides 15 cm and 18 cm. The height corresponding to a 15-cm base is 9 cm. Find the area of the parallelogram. the product of Abase of a parallelogram is any of its sides. The 15 Quick Check. half and the corresponding A Substitute 9 for b and Simplify. 䉭XYZ has area © Pearson Education, Inc., publishing as Pearson Prentice Hall. base 135 ) b Theorem 10-3: Area of a Triangle a A 15 2 Finding the Area of a Triangle Find thearea of 䉭XYZ. h . bh The area of a triangle is ( A5 The area of a parallelogram is the product of a base All rights reserved. height All rights reserved. and A Area of a parallelogram A 9 The area of the parallelogram is The area of a rectangle is the product of 119 Examples. Theorem 10-1: Area of aRectangle base Geometry Lesson 9-7 1 Finding the Area of a Parallelogram A parallelogram has 9-in. and 18-in. sides. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes Vocabulary and Key Concepts. its Daily Notetaking Guide Name_____________________________________Class____________________________ Date ________________ Areas of Parallelograms and Triangles Lesson Objectives 1 Find the area of a parallelogram 2 Find the area of a triangle line symmetry, rotational symmetry, glide reflectional symmetry, translational symmetry L1Name_____________________________________ Class____________________________ Date________________ Lesson 10-1 translation 2. List the symmetries in the tessellation. © Pearson Education, Inc., publishing as Pearson Prentice Hall. A repeated combination of an All rights reserved. Glidereflectional symmetry is the type of symmetry for which there is a glide reflection that 135 cm2 2. Find the area of the triangle. 5 cm 30 cm2 13 cm 12 cm The height of a triangle is the length of the altitude to the line h containing that base. b 120 30 Geometry Lesson 10-1 Geometry Daily Notetaking GuideL1 L1 Daily Notetaking Guide Geometry Lesson 10-1 All-In-One Answers Version B 121 L1 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 10-2 1 Find the area of atrapezoid Find the area of a rhombus or a kite R 2 Finding the Area of a Rhombus Find the area of rhombus RSTU. Areas of Trapezoids, Rhombuses, and Kites Lesson Objectives 2 Name_____________________________________ Class____________________________ Date ________________NAEP 2005 Strand: Measurement To find the area, you need to know the lengths of both diagonals. Draw diagonal SU, and label the intersection of the diagonals point X. Topic: Measuring Physical Attributes 䉭SXT is a Local Standards: ____________________________________ are perpendicular.right S 24 triangle because the diagonals of a rhombus The diagonals of a rhombus bisect each other, so TX 12 ft 13 ft X U T ft. You can use the Pythagorean triple 5, 12, 13 or the Pythagorean Theorem Vocabulary and Key Concepts. to conclude that SX SU The area of a trapezoid is half the product ofthe height and the sum of A 5 12 A 5 12 All rights reserved. b1 the bases. h 1h(b 1 b ) 1 2 2 A b2 Theorem 10-5: Area of a Rhombus or a Kite The area of a rhombus or a kite is half the product of the lengths of its d2 d1 diagonals. All rights reserved. Theorem 10-4: Area of a Trapezoid d1 ( 5 ft. ft becausethe diagonals of a rhombus bisect each other. 24 d2 )( Area of a rhombus ) 10 120 24 Substitute for d1 and 10 for d2 . Simplify. ft 2. 120 The area of rhombus RSTU is Quick Check. 1d d 2 1 2 A A5 10 1. Find the area of a trapezoid with height 7 cm and bases 12 cm and 15 cm. 94.5 cm 2 The height of atrapezoid is the perpendicular Leg h All rights reserved. b2 Base Examples. 1 Applying the Area of a Trapezoid A car window is 20 in. shaped like the trapezoid shown. Find the area of the window. 18 in. 36 in. A 5 12 h A 5 12 ( A5 ( b1 )( 18 ) b2 20 36 ) 18 Substitute 504 5 2 ⫹ 12 2 ⫽ 13 2 for h, 20 for b1 ,and 36 for b2 . Simplify. in.2. 504 Geometry Lesson 10-2 Daily Notetaking Guide L1 L1 Daily Notetaking Guide Name_____________________________________ Class____________________________ Date ________________ Lesson 10-3 2 .Finding the Area of a Regular Polygon A library is in theshape Areas of Regular Polygons a2 a p 9.0 1 ap 2 81 )2 ( a2 All rights reserved. perimeter. All rights reserved. The area of a regular polygon is half the product of the apothem and the A ( a2 Theorem 10-6: Area of a Regular Polygon a< of a regular polygon is the distance from the center to a vertex. 18.0A 12 ap 21.7 A 12 1562.4 A Examples. 1 Finding Angle Measures This regular hexagon has an apothem and radii drawn. Find the measure of each numbered angle. m⬔2 1 2 1 2 3 Divide 360 by the number of sides. The apothem bisects the vertex angle of m⬔1 isosceles the triangle formed by the radii.m⬔2 1 2 ( 60 ) 90 m⬔3 180 ( m⬔1 , m⬔2 124 60 Geometry Lesson 10-3 30 30 Substitute ) 60 30 , and m⬔3 60 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. Apothem 6 471.25 21.7 ) 144 )( Substitute 8 for n and ) 144To the nearest 10 ft 2, the area is Radius 60 9.0 ft Solve for a. 18.0 for s, and simplify. Step 3 Find the area A. ( 9.0 ft Pythagorean Theorem Find the perimeter, where n ⴝ the number of sides of a regular polygon. ( Center 360 )2 p ns p (8) The apothem of a regular polygon is the perpendicular distancefrom the center to a side. m⬔1 23.5 552.25 Step 2 Find the perimeter p. The center of a regular polygon is the center of the circumscribed circle. radius 23.5 ft Step 1 Find the apothem a. Vocabulary and Key Concepts. The 18.0 ft of a regular octagon. Each side is 18.0 ft. The radius of the octagon is 23.5ft. Find the area of the library to the nearest 10 ft 2. Consecutive radii form an isosceles triangle, so an apothem bisects the side of the octagon. To apply the area formula A 12ap, you need to find a and p. NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes Local Standards:____________________________________ L1 123 Geometry Lesson 10-2 Name_____________________________________ Class____________________________ Date________________ Lesson Objective 1 Find the area of a regular polygon © Pearson Education, Inc., publishing as PearsonPrentice Hall. 2. Critical Thinking In Example 2, explain how you can use a Pythagorean triple to conclude that XU 5 ft. Area of a trapezoid The area of the car window is 122 © Pearson Education, Inc., publishing as Pearson Prentice Hall. b1 Leg © Pearson Education, Inc., publishing as Pearson PrenticeHall. Base distance h between the bases. Area of a regular Substitute 21.7 polygon for a and 144 for p. Simplify. 1560 ft 2. Quick Check. 1. At the right, a portion of a regular octagon has radii and an apothem drawn. Find the measure of each numbered angle. ml1 5 45; ml2 5 22.5; ml3 5 67.5 1 2 3 2. Findthe area of a regular pentagon with 11.6-cm sides and an 8-cm apothem. 232 cm 2 for m⬔1. The sum of the measures of the angles of a triangle is 180. 60 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 10-3 Geometry 125 31 Geometry: All-In-OneAnswers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Name_____________________________________ Class____________________________ Date ________________ Lesson 10-4 3 Using SimilarityRatios Benita plants the same crop in two rectangular Perimeters and Areas of Similar Figures Lesson Objective 1 fields. Each dimension of the larger field is 3 12 times the dimension of the smaller field. Seeding the smaller field costs $8. How much money does seeding the larger field cost? NAEP 2005Strand: Measurement and Number Properties and Operations Find the perimeters and areas of similar figures Topics: Systems of Measurement; Ratios and Proportional Reasoning The similarity ratio of the fields is 3.5 : 1, so the ratio of the areas of the . fields is (3.5) 2 : (1) 2, or 12.25 to 1 LocalStandards: ____________________________________ Because seeding the smaller field costs $8, seeding 12.25 times as much land costs $8 . 12.25 Key Concepts. All rights reserved. If the similarity ratio of two similar figures is ba , then a b (1) the ratio of their perimeters is a2 b2 (2) the ratio of theirareas is and All rights reserved. Seeding the larger field costs Theorem 10-7: Perimeters and Areas of Similar Figures . $98 . Quick Check. 1. Two similar polygons have corresponding sides in the ratio 5 : 7. a. Find the ratio of their perimeters. b. Find the ratio of their areas. 5:7 25 : 49 Examples. 5 theshortest side of the right-hand triangle has length 5 4 From larger to smaller, the similarity ratio is 6 5 6.25 5 , and 7.5 . . By the Perimeters and Areas of Similar Figures Theorem, the ratio of the perimeters is 5 4 52 42 , and the ratio of the areas is , or 25 16 . 2 Finding Areas Using Similar Figures The ratioof the length of the corresponding sides of two regular octagons is 83. The area of the larger octagon is 320 ft 2. Find the area of the smaller octagon. All regular octagons are similar. Because the ratio of the lengths of the corresponding sides of the regular 82 32 octagons is 83, the ratio of their areas is320 A 9 64 A A 64 9 . 3. The similarity ratio of the dimensions of two similar pieces of window glass is 3 : 5. The smaller piece costs $2.50. What should be the cost of the larger piece? $6.94 Write a proportion. 2880 45 Use the Cross-Product Property. 64 . Divide each side by The area of the smalleroctagon is 126 , or 96 in.2 ft 2. 45 Geometry Lesson 10-4 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Lesson 10-5 Geometry Lesson 10-4 127 Name_____________________________________Class____________________________ Date ________________ A 12 Trigonometry and Area A NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes Lesson Objectives 1 Find the area of a regular polygon using trigonometry 2 Find the area of a triangle using trigonometry DailyNotetaking Guide 180(sin 36ⴗ) Substitute for a and p. (cos 36) (sin 36) Simplify. 770.355778 A The area is about Local Standards: ____________________________________ 18(cos 36ⴗ) 1620 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 64 2. The corresponding sides of two similarparallelograms are in the ratio 34. The area of the smaller parallelogram is 54 in.2. Find the area of the larger parallelogram. All rights reserved. 4 The shortest side of the left-hand triangle has length © Pearson Education, Inc., publishing as Pearson Prentice Hall. 4 similar. Find the ratio (larger to smaller)of their perimeters and of their areas. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 Finding Ratios in Similar Figures The triangles at the right are Use a calculator. 770 ft2. 2 Surveying A triangular park has two sides that measure 200 ft and 300 ft Key Concepts. and form a 65 angle.Find the area of the park to the nearest hundred square feet. Theorem 10-8: Area of a Triangle Given SAS c . A 1 bc(sin A) 2 Area of 䉭ABC a C b All rights reserved. the sine of the included angle B and the lengths of two sides All rights reserved. Use Theorem 10-8: The area of a triangle is The area of atriangle is one half the product of product of the lengths of two sides and the one half sine the 300 ft of the included angle. Area 12 side length side length sine of included angle Theorem 10-8 Area 12 Substitute. Area Examples. 200 ft 200 30,000 300 sin 65ⴗ sin 65ⴗ Substitute. 27189.23361 The area ofthe park is approximately 65 Use a calculator. 27,200 ft2. 1 Finding Area The radius of a garden in the shape of a regular pentagon formula A 12 ap. C 18 ft 360 Because a pentagon has five sides, m⬔ACB 5 72 . A M B CA and CB are radii, so CA CB . Therefore, 䉭ACM 䉭BCM by the HL Theorem, som⬔ACM 12m⬔ACB Use the cosine ratio to find a. cos 36 a 18 a 18(cos 36ⴗ) 36 . Use the sine ratio to find AM. sin Use the ratio. 36 AM 18 AM Solve. 18(sin 36ⴗ) Use AM to find p. Because 䉭ACM 䉭BCM, AB 2 ? AM . Because the pentagon is regular, p 5 ? AB . So p 5(2 ? AM ) 10 ? AM 10 ? 18(sin36ⴗ) © Pearson Education, Inc., publishing as Pearson Prentice Hall. Find the perimeter p and apothem a, and then find the area using the © Pearson Education, Inc., publishing as Pearson Prentice Hall. is 18 feet. Find the area of the garden. Quick Check. 1. Find the area of a regular octagon with aperimeter of 80 in. Give the area to the nearest tenth. 482.8 in.2 2. Two sides of a triangular building plot are 120 ft and 85 ft long. They include an angle of 85. Find the area of the building plot to the nearest square foot. 5081 ft 2 180(sin 36ⴗ) . Finally, substitute into the area formula A 12ap. 128 32Geometry Lesson 10-5 Geometry Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 10-5 All-In-One Answers Version B 129 L1 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________Date________________ Lesson 10-6 Topic: Measuring Physical Attributes 0 0 0 mXY mXD mDY 0 0 mXY m/ XCD mDY 0 mXY 56 40 0 mXY 96 Local Standards: ____________________________________ C B measures of the two arcs. 1 0 0 mABC mAB 1 mBC A Theorem 10-9: Circumference of aCircle p times the diameter d r O 2pr C Theorem 10-10: Arc Length The length of an arc of a circle is the product of the ratio measure of the arc and the A circumference of the circle . O 0 mAB ? 2pr 360 All rights reserved. A circle is the set of all points equidistant from a given point called the center. A of acircle is the point from which all points are equidistant. center A radius is a segment that has one endpoint at the center and the other endpoint on the circle. A is an angle whose vertex is the center of the circle. central angle Geometry Lesson 10-6 Daily Notetaking Guide D Arc Addition PostulateSubstitute. 56 C W X Simplify. 2π 360 ( ) 18 B 150 18 cm M D A Substitute. Quick Check. 1 0 1 1. Use the diagram in Example 1. Find m⬔YCD, mYMW , mMW, and mXMY. 40; 180; 96; 264 2. Find the length of a semicircle with radius 1.3 m in terms of π. 1.3p m L1 Daily Notetaking Guide GeometryLesson 10-6 131 Name_____________________________________ Class____________________________ Date ________________ Lesson 10-7 2 Finding the Area of a Sector of a Circle Find the area of sector ACB. Areas of Circles and Sectors Leave your answer in terms of π. NAEP 2005 Strand:Measurement Topic: Measuring Physical Attributes area of sector ACB mAB Local Standards: ____________________________________ Vocabulary and Key Concepts. Theorem 10-11: Area of a Circle measure of the arc The area of a sector of a circle is the product of the ratio 360 . area of the circle.A r O B 5 18 The area of sector ACB is )2 π( 6 360 10 6 m 100 36 π π 10π B A Substitute. Simplify. Simplify. m 2. about 41 in.2 Segment intercepted arc. © Pearson Education, Inc., publishing as Pearson Prentice Hall. of a circle is the part bounded by an arc and Examples. Sector 1 Applying the Area ofa Circle A circular archery target has a 2-ft diameter. of a circle It is yellow except for a red bull’s-eye at the center with a 6-in. diameter. Find the area of the yellow region to the nearest whole number. First find the areas of the archery target and the red bull’s-eye. The radius of the archery target is 12 2 ft1 ft 12 in. ( The area of the archery target is πr 2 ) π ( ) 2 12 (6 π( )2 The area of the red region is area of archery target πr 2 in. 3 9p 9p 135 424.11501 The area of the yellow region is about Geometry Lesson 10-7 3 in. in.2 area of red region 144p ) The radius of the red bull’s-eye region is 12 in.2 ©Pearson Education, Inc., publishing as Pearson Prentice Hall. of a circle segment the segment joining its endpoints. 132 100 C πr 2 1. How much more pizza is in a 14-in.-diameter pizza than in a 12-in. pizza? A sector of a circle is a region bounded by two radii and their 144p 5 360 Quick Check. 0 mAB ?pr2 360 Area of sector AOB A r O Theorem 10-12: Area of a Sector of a Circle and the . pr 2 All rights reserved. A All rights reserved. the product of p and the square of the radius The area of a circle is L1 Y 40 Simplify. Name_____________________________________Class____________________________ Date________________ Lesson Objective 1 Find the areas of circles, sectors, and segments of circles © Pearson Education, Inc., publishing as Pearson Prentice Hall. L1 Substitute. 21 π 1 The length of ADB is 21π cm. Circumference of a circle is the distancearound the circle. Pi () is the ratio of the circumference of a circle to its diameter. 130 210 M The measure of a minor arc is the measure of its corresponding central angle. 1 2 Finding Arc Length Find the length of ADB in circle M in terms of π. 0 Because mAB 150, 1 . Arc Addition Postulate mADB 360150 210 1 1 length of ADB mADB Arc Length Postulate 360 ? 2 pr B 360 0 length of AB 1 Arc Addition Postulate 0 1 1 mDXM mDX mXWM 1 mDXM 56 180 1 236 mDXM r © Pearson Education, Inc., publishing as Pearson Prentice Hall. or C pd . © Pearson Education, Inc., publishing as PearsonPrentice Hall. C All rights reserved. Postulate 10-1: Arc Addition Postulate The measure of the arc formed by two adjacent arcs is the sum of the All rights reserved. Vocabulary and Key Concepts. The circumference of a circle is 0 1 Finding the Measures of Arcs Find mXY and mDXM in circle C. NAEP2005 Strand: Measurement Find the measures of central angles and arcs Find circumference and arc length 2 Examples. Circles and Arcs Lesson Objectives 1 Name_____________________________________ Class____________________________ Date ________________ 2. Critical Thinking Acircle has a diameter of 20 cm. What is the area of a sector bounded by a 208 major arc? Round your answer to the nearest tenth. 181.5 cm 2 area of yellow region 424 π Simplify. in.2 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 10-7 Geometry133 33 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date________________ Lesson 10-8 Example. Geometric Probability 2 Finding Probability Using Area A circle is inscribed in a square target NAEP2005 Strand: Data Analysis and Probability Use segment and area models to find the probabilities of events Local Standards: ____________________________________ Find the area of the square. A s2 Vocabulary. 20 2 400 cm 2 Find the area of the circle. Because the square has sides of length 20cm, the circle’s diameter is 20 cm, so its radius is 10 cm. Example. 1 Finding Probability Using Segments A gnat lands at random on the A πr 2 π All rights reserved. All rights reserved. Geometric probability is a model in which you let points represent outcomes. ( 10 )2 cm 2 100π Find the area of theregion between the square and the circle. ( A 400 100π ) cm 2 Use areas to calculate the probability that a dart landing randomly in the square does not land within the circle. Use a calculator. Round to the nearest thousandth. area between square and circle P(between square and circle) area of squareedge of the ruler below. Find the probability that the gnat lands on a point between 2 and 10. 400 ⴚ 100π 3 4 5 6 7 8 9 The length of the segment between 2 and 10 is 12 The length of the ruler is 10 10 11 400 2 8 . . P(landing between 2 and 10) length of favorable segment length of entire segment 8 1223 Quick Check. 1. A point on AB is selected at random. What is the probability that it is a point on CD? A 1 2 3 4 5 6 7 D B 8 9 10 1 4 0.215 The probability that a dart landing randomly in the square does not land 21.5 within the circle is about %. Quick Check. 2. Use the diagram in Example 2. If youchange the radius of the circle as indicated, what then is the probability of hitting outside the circle? a. Divide the radius by 2. about 80.4% b. Divide the radius by 5. 2 5 134 π about 96.9% Geometry Lesson 10-8 Daily Notetaking Guide L1 L1 Name_____________________________________Class____________________________ Date ________________ Lesson 11-1 Geometry Lesson 10-8 135 Name_____________________________________ Class____________________________ Date ________________ 2 Using Euler’s Formula Use Euler’s Formula to find the number of SpaceFigures and Cross Sections Lesson Objective 1 Recognize polyhedra and their parts 2 Visualize cross sections of space figures Daily Notetaking Guide edges on a solid with 6 faces and 8 vertices. FVE2 NAEP 2005 Strand: Geometry Topic: Dimension and Shape 6 8 E 2 12 Local Standards:____________________________________ E Euler’s Formula © Pearson Education, Inc., publishing as Pearson Prentice Hall. 0 C © Pearson Education, Inc., publishing as Pearson Prentice Hall. 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 20 cm with 20-cm sides. Find theprobability that a dart landing randomly within the square does not land within the circle. Topic: Probability All rights reserved. Lesson Objective 1 Name_____________________________________ Class____________________________ Date ________________ Substitute the number of faces andvertices. Simplify. A solid with 6 faces and 8 vertices has 12 edges. Vocabulary and Key Concepts. Quick Check. Euler’s Formula F⫹V⫽E⫹2 . A polyhedron is a three-dimensional figure whose surfaces are polygons. A face 1. List the vertices, edges, and faces of the polyhedron. All rights reserved. arerelated by the formula All rights reserved. The numbers of faces (F ), vertices (V), and edges (E) of a polyhedron Rectangular prism R S T U 2. Use Euler’s Formula to find the number of edges on a polyhedron with eight triangular faces. is a flat surface of a polyhedron in the shape of a polygon. R, S, T, U,V; RS, RU, RT, VS, VU, VT, SU, UT, TS; kRSU, kRUT, kRTS, kVSU, kVUT, kVTS V 12 edges An edge is a segment that is formed by the intersection of two Faces faces. is a point where three or more edges intersect. Vertex Examples. 1 Identifying Vertices, Edges, and Faces How many vertices, edges,and faces are there in the polyhedron shown? Give a list of each. There are four There are six There are four vertices: B A . A, B, C, and D C 136 34 edges: AB, BC, CD, DA, AC, and BD . D © Pearson Education, Inc., publishing as Pearson Prentice Hall. vertex A cross section is the intersection of a solidand a plane. © Pearson Education, Inc., publishing as Pearson Prentice Hall. A Edge faces: kABC, kABD, kACD, and kBCD . Geometry Lesson 11-1 Geometry Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 11-1 All-In-One Answers Version B 137 L1 Geometry: All-In-OneAnswers Version B (continued) Name_____________________________________ Class____________________________ Date ________________ Lesson 11-2 Name_____________________________________ Class____________________________ Date________________ A cylinder SurfaceAreas of Prisms and Cylinders is a three-dimensional figure with two congruent circular bases that lie in parallel planes. Lesson Objectives 1 Find the surface area of a prism Find the surface area of a cylinder 2 In a cylinder, the bases are parallel circles. Topic: Measuring Physical Attributes h LocalStandards: ____________________________________ The . height of the cylinder L.A. 2πrh, or L.A. πdh The surface area of a right cylinder is the sum of the areas of the two circular bases . Area 5 2prh 1 2 r r Area of a base B π r2 prism Pentagonal faces. of a polyhedron are the parallel bases faces.Bases Lateral edges Lateral faces are the faces on a prism that are not bases. The of a prism is the sum of the areas lateral area of the lateral faces. Lateral faces The surface area of a prism or a cylinder is the sum of the p Substitute the formulas for lateral area of a cylinder and area of a circle. r2 Baseslateral area and the areas of the two bases. 5 24 p 1 8 5 32 p S.A. 5 L.A. 1 2 Use the formula for surface area. ( )( 6 ) 2π ( 2 2) A prism is a polyhedron with exactly two congruent, parallel The Barley Container B 2π 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Surface Area 2πr Allrights reserved. h Lateral Find the surface area of each container. Remember that r d2. S.A. 5 L.A. 1 2 r h Finding Surface Area of Cylinders A company sells cornmeal and barley in cylindrical containers. The diameter of the base of the 6-in.-high cornmeal container is 4 in. The diameter of the base of the4-in.-high barley container is 6 in. Which container has the greater surface area? Cornmeal Container 2πr 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. S.A. L.A. 2B, or S.A. 2πrh lateral area All rights reserved. All rights reserved. and the cylinders right Example. Theorem 11-1:Lateral and Surface Area of a Cylinder The lateral area of a right cylinder is the product of the and the of a cylinder is the area of the curved lateral area surface. Vocabulary and Key Concepts. circumference of the base h Bases NAEP 2005 Strand: Measurement 5 2 p 2π Substitute for r and h. B h 2pr2 (3 )( 4 ) 2π ( 3 2) Simplify. 5 24 p 1 Combine like terms. 5 42 p p Because 42π in.2 ⬎ 32π in.2, the surface area. r barley 18 p container has the greater Quick Check. a. Find the surface area of a cylinder with height 10 cm and radius 10 cm in terms of π. 400p cm2 b. The company in the Example wants tomake a label to cover the cornmeal container. The label will cover the container all the way around, but will not cover any part of the top or bottom. What is the area of the label to the nearest tenth of a square inch? 75.4 in.2 prism Triangular 138 Geometry Lesson 11-2 Daily Notetaking Guide L1 L1Name_____________________________________ Class____________________________ Date________________ Lesson 11-3 © Pearson Education, Inc., publishing as Pearson Prentice Hall. The perimeter p of the square base is 4 You are given ᐍ 12 find the lateral area. NAEP 2005 Strand:Measurement Topic: Measuring Physical Attributes Local Standards: ____________________________________ L.A. 5 12 slant height ᐉ . 1 pᐉ 2 L.A. B The surface area of a regular pyramid is the sum of the lateral area S.A. L.A. and the area of the base All rights reserved. and the ( 30 )( 7.5 ft, or ftand you found that p ᐍ p 2 L.A. 5 All rights reserved. perimeter of the base 1 L.A. 5 Vocabulary and Key Concepts. Theorem 11-3: Lateral and Surface Area of a Regular Pyramid The lateral area of a regular pyramid is half the product of the Geometry Lesson 11-2 139Name_____________________________________ Class____________________________ Date ________________ Surface Areas of Pyramids and Cones Lesson Objectives 1 Find the surface area of a pyramid 2 Find the surface area of a cone Daily Notetaking Guide 30 30 ft. ft, so you can Use theformula for lateral area of a pyramid. ) 12 180 Substitute. Simplify. Find the area of the square base. Because the base is a square with side length 7.5 ft, B s2 7.5 2 56.25 S.A. 5 L.A. 1 B . S.A. 5 B S.A. 5 A regular pyramid is a pyramid whose base is a regular polygon and whose lateral faces are 180 1 .Use the formula for surface area of a pyramid. 56.25 236.25 Substitute. Simplify. The surface area of the square pyramid is 236.25 ft2. congruent isosceles triangles. vertex to the plane of the base. The height of a pyramid or a cone is the length of Lateral Vertex edge Lateral face the altitude. The Altitudeof a regular pyramid slant height Base is the length of the altitude of a lateral face. The lateral area of a pyramid is is the sum of the Base edge Regular Pyramid areas of the congruent lateral faces. The of a pyramid is the sum of the lateral area and the area of the base. surface area Example. FindingSurface Area of a Pyramid Find the surface area of a square pyramid with base edges 7.5 ft and slant height 12 ft. 12 ft © Pearson Education, Inc., publishing as Pearson Prentice Hall. of a pyramid or cone is the perpendicular segment from the altitude © Pearson Education, Inc., publishing as PearsonPrentice Hall. The Quick Check. Find the surface area of a square pyramid with base edges 5 m and slant height 3 m. 55 m 2 7.5 ft 140 L1 Geometry Lesson 11-3 Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 11-3 Geometry 141 35 Geometry: All-In-One Answers Version B (continued) Name_____________________________________ Class____________________________ Date ________________ 2 Finding Volume of a Triangular Prism Find the volume of the prism. Lesson 11-4 Volumes of Prisms and Cylinders Lesson Objectives Find thevolume of a prism Find the volume of a cylinder Topic: Measuring Physical Attributes is the Local Standards: ____________________________________ Pythagorean Theorem to calculate the length of the other leg. height of the prism V h 20 2 841 )( h 400 40 m . Use the ) 20 m 210 441 21 . Use theUse the formula for the volume of a prism. V 210 ? V 8400 40 Substitute. Simplify. 8400 The volume of the triangular prism is B m3. Bh Quick Check. h 1. The cylinder shown is oblique. a. Find its volume in terms of π. r B pr 2h Volume is the space that a figure occupies. A space figure is a three-dimensional figure that is the combination composite of two or more simpler figures. Examples. 1 Finding Volume of a Cylinder Find the volume of the cylinder. Leave your answer in terms of π. The formula for the volume of a cylinder is The diagram shows h and d, but you must find r. V ⴝ pr 2h . 9 ft r 512d 5 8 V5 p r2 h 5p? 82 ? 9 5 576 16 ft Use the formula for the volume of a cylinder. 8m b. Find its volume to the nearest tenth of a cubic meter. 804.2 m 3 2. Find the volume of the triangular prism. 150 m 3 10 m 5m Substitute. 6m p Simplify. The volume of the cylinder is ft3. 576π Geometry Lesson 11-4Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Lesson 11-5 Daily Notetaking Guide of the oblique cone in terms of π. 1 r5 d 5 3 in. 2 11 in. Local Standards: ____________________________________Key Concepts. height of the pyramid and the . 1Bh 3 h B Theorem 11-9: Volume of a Cone All rights reserved. 5 All rights reserved. area of the base and the h 33 Use the formula for the volume of a cone. 11 ) p Substitute 3 for r and 11 6 in. for h. Simplify. The volume of the cone is 33p in.3. Quick Check.1. Find the volume of a square pyramid with base edges 24 m and slant height 13 m. 960 m 3 The volume of a cone is one third the product of the area of the base r2 5 13 p( 3 2)( Theorem 11-8: Volume of a Pyramid The volume of a pyramid is one third the product of the 143 2 Finding Volume of anOblique Cone Find the volume NAEP 2005 Strand: Measurement Topic: Measuring Physical Attributes V 5 13 p height of the cone . h 1pr 2h 3 V 13Bh, or V Geometry Lesson 11-4 Name_____________________________________ Class____________________________ Date ________________Volumes of Pyramids and Cones Lesson Objectives 1 Find the volume of a pyramid 2 Find the volume of the cone V 16 m 256p m3 All rights reserved. V Bh, or V area of a base . © Pearson Education, Inc., publishing as Pearson Prentice Hall. height of the cylinder and the © Pearson Education, Inc.,publishing as Pearson Prentice Hall. Theorem 11-7: Volume of a Cylinder The volume of a cylinder is the product of the 142 V B All rights reserved. All rights reserved. area of a base . ( . volume Theorem 11-6: Volume of a Prism The volume of a prism is the product of the 2 29 where one leg altitude andthe other leg is the 21 The area B of the base is 21bh 12 20 area of the base to find the volume of the prism. Theorem 11-5: Cavalieri’s Principle If two space figures have the same height and the same cross sectional area at every level, then they have the same base 29 m with triangular bases. The righttriangle base of the triangular prism is a Vocabulary and Key Concepts. and the right triangular prism The prism is a NAEP 2005 Strand: Measurement © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 2 Name_____________________________________Class____________________________ Date________________ r Examples. 1 Finding Volume of a Pyramid Find the volume of a square pyramid with base edges 15 cm and height 22 cm. Because the base is a square, B V 5 13 B 13 V5 ( h 225 )( 22 1650 ) 36 ? 15 . 225 Substitute 225 for B and 22 forh. Simplify. The volume of the square pyramid is 144 15 Use the formula for volume of a pyramid. Geometry Lesson 11-5 Geometry 1650 cm3. Daily Notetaking Guide L1 © Pearson Education, Inc., publishing as Pearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. B 2.A small child’s teepee is in the shape of a cone 6 ft tall and 7 ft in diameter. Find the volume of the teepee to the nearest cubic foot. 77 ft 3 L1 Daily Notetaking Guide Geometry Lesson 11-5 All-In-One Answers Version B 145 L1 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ Date ________________ Lesson 11-6 Examples. Surface Areas and Volumes of Spheres Lesson Objective 1 Name_____________________________________Class____________________________ Date________________ 1 Finding Surface Area The circumference of a rubber ball is 13 cm. NAEP 2005 Strand: Measurement Find the surface area and volume of a sphere Calculate its surface area to the nearest whole number. Topic: Measuring PhysicalAttributes Step 1 First, find the radius. Local Standards: ____________________________________ C 5 2pr 13 Vocabulary and Key Concepts. 13 Theorem 11-10: Surface Area of a Sphere 2p square S.A. of the radius r All rights reserved. the of the sphere. 4pr 2 Theorem 11-11: Volume of a SphereSubstitute 5r Solve for r. S.A. 5 4pr2 5 4p ? The volume of a sphere is four thirds the product of p and the of the cube radius of the sphere. for C. Use the formula for surface area of a sphere. ( 13 2p ) 2 13 Substitute for r. 2p 169 5 r 4pr 3 3 S.A. 13 Step 2 Use the radius to find the surface area. All rightsreserved. The surface area of a sphere is four times the product of p and Use the formula for circumference. 5 2pr Simplify. p 53.794371 < Use a calculator. To the nearest whole number, the surface area of the rubber ball is All rights reserved. diameter of a sphere is the given point from which all pointson the sphere are equidistant. r The radius of a sphere is a segment that has one endpoint at the center and the other endpoint on the sphere. diameter The d of a sphere is a segment passing circumference through the center with endpoints on the sphere. sphere A great circle is the intersection of asphere and a plane containing the center of the sphere. great circle The divides a sphere into two Leave your answer in terms of π. V 5 34 p 5 15 4500 Use the formula for volume of a sphere. Substitute r ⴝ 30 ⴝ 2 3 p 15 . Simplify. 4500p The volume of the sphere is cm3. Quick Check. 1. Find the surfacearea of a sphere with d 14 in. Give your answer in two ways, in terms of π and rounded to the nearest square inch. 196p in.2, 616 in.2 Great 2. Find the volume to the nearest cubic inch of a sphere with diameter 60 in. circle The circumference of a sphere is the circumference of its 113,097 in.3 greatcircle. Geometry Lesson 11-6 Daily Notetaking Guide L1 Daily Notetaking Guide L1 Name_____________________________________ Class____________________________ Date________________ Geometry Lesson 11-6 147 Name_____________________________________Class____________________________ Date ________________ 2 Finding the Similarity Ratio Find the similarity ratio of two similar Lesson 11-7 cylinders with surface areas of 98π ft 2 and 2π ft 2. Use the ratio of the surface areas to find the similarity ratios. Areas and Volumes of Similar Solids LessonObjective 1 Find relationships between the ratios of the areas and volumes of similar solids © Pearson Education, Inc., publishing as Pearson Prentice Hall. r3 5 43p ? Hemispheres congruent hemispheres. 146 cm2. m 30 c center © Pearson Education, Inc., publishing as Pearson Prentice Hall. centergiven point. © Pearson Education, Inc., publishing as Pearson Prentice Hall. A sphere is the set of all points in space equidistant from a The 54 2 Finding Volume Find the volume of the sphere. NAEP 2005 Strand: Measurement Topic: Systems of Measurement a2 2 Vocabulary and Key Concepts. bTheorem 11-12: Areas and Volumes of Similar Solids If the similarity ratio of two similar solids is a : b, then 5 (2) the ratio of their volumes is a3 : b3 . All rights reserved. All rights reserved. a b5 (1) the ratio of their corresponding areas is a2 : b2 , and 98p a2 5 b2 Local Standards:____________________________________ The ratio of the surface areas is a2 : b2 . 2p 49 Simplify. 1 7 Take the square root of each side. 1 The similarity ratio is 7 : 1 . Quick Check. 1. Are the two cylinders similar? If so, give the similarity ratio. Similar solids have the same shape and all of theircorresponding parts are proportional. yes; 6 : 5 The 10 12 of two similar objects is the ratio of their corresponding linear similarity ratio 6 © Pearson Education, Inc., publishing as Pearson Prentice Hall. Examples. 1 Identifying Similar Solids Are the two solids similar? If so, give the similarity ratio. 26 in. 8in. 9 in. 3 in. © Pearson Education, Inc., publishing as Pearson Prentice Hall. dimensions. 5 2. Find the similarity ratio of two similar prisms with surface areas 144 m 2 and 324 m 2. 2:3 Both figures have the same shape. Check that the ratios of the corresponding dimensions are equal. The ratio of the radiiis The cones are 148 L1 3 9 not similar Geometry Lesson 11-7 , and the ratio of the heights is because 8 3 9 ⬆ 26 8 26 . . Daily Notetaking Guide All-In-One Answers Version B L1 L1 Daily Notetaking Guide Geometry Lesson 11-7 Geometry 149 37 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ Lesson 12-1 Applying Tangent Lines A belt fits tightly around two circular pulleys, as shown. Find the distance between the centers of the pulleys. Round your answer to the nearesttenth. NAEP 2005 Strand: Geometry Use the relationship between a radius and a tangent Use the relationship between two tangents from one point 2 Example. Tangent Lines Lesson Objectives 1 Name_____________________________________ Class____________________________ Date________________ Topic: Relationships Among Geometric Figures Local Standards: ____________________________________ Y O Vocabulary and Key Concepts. P D Z 15 cm Because opposite sides of a rectangle have the same measure, DW 3 cm and OD A If a line is tangent to a circle, then theline is perpendicular to the radius drawn P to the point of tangency. * ) B O AB ' OP All rights reserved. All rights reserved. Theorem 12-1 Theorem 12-2 the line is tangent to the circle . AB is tangent to 䉺 O. right of a supplement right Because the radius of 䉺 P is 7 cm, PD 7⫺3⫽4 OD2 PD2 OP2 2 angle,triangle. cm. Pythagorean Theorem 4 2 OP2 Substitute. OP2 241 W cm. ⬔ODP is also a right angle, and 䉭OPD is a B O 15 Because ⬔ODP is the 15 A P If a line in the plane of a circle is perpendicular to a radius at its endpoint * ) 7 cm 3 cm Draw OP. Then draw OD parallel to ZW to form rectangleODWZ. Because OZ is a radius of 䉺 O, OZ 3 cm. on the circle, then 15 cm X Simplify. 15.524175 OP Use a calculator to find the square root. The distance between the centers of the pulleys is about 15.5 cm. congruent B . O AB > CB C A tangent to a circle is a line, segment, or ray in the plane of thecircle that intersects the circle in exactly one point. The point of tangency is the point where a circle and a tangent intersect. A tangent B point of tangency Quick Check. A belt fits tightly around two circular pulleys, as shown. Find the distance between the centers of the pulleys. All rights reserved. are A ©Pearson Education, Inc., publishing as Pearson Prentice Hall. The two segments tangent to a circle from a point outside the circle © Pearson Education, Inc., publishing as Pearson Prentice Hall. Theorem 12-3 35 in. 14 in. 8 in. about 35.5 in. A triangle is inscribed in a circle if all vertices of the triangle lieon the circle. A triangle is circumscribed about a circle if each side of the triangle is tangent to the circle. Geometry Lesson 12-1 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Lesson 12-2 ⬔AOX AXBX lBOX congruent All rights reserved. congruent congruent chords. arcs. central angles. All rights reserved. congruent Theorem 12-4 Within a circle or in congruent circles (3) Congruent arcs have (1) Chords equidistant from the center are equidistant . congruent congruent central angles have X Ochords. congruent chords have arcs. B B 2 Using Theorem 12-5 Find AB. Theorem 12-5 Within a circle or in congruent circles (2) Congruent chords are by the definition of an angle bisector. because 0 0 AX BX because (2) Congruent chords have A What can you conclude? Local Standards:____________________________________ congruent 151 Examples. 1 Using Theorem 12-4 In the diagram, radius OX bisects ⬔AOB. NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures Vocabulary and Key Concepts. (1) Congruent central angles have Geometry Lesson 12-1 Name_____________________________________ Class____________________________ Date ________________ Chords and Arcs Lesson Objectives 1 Use congruent chords, arcs, and central angles 2 Recognize properties of lines through the center of a circle Daily Notetaking Guide © PearsonEducation, Inc., publishing as Pearson Prentice Hall. 150 QS QR RS Segment Addition Postulate QS 7 7 Substitute. QS 14 Simplify. AB QS Chords that are equidistant from the center of a circle are congruent. AB 14 Substitute 4 C 4 A Q 7 14 R 7 S for QS. from the center. Theorem 12-7 In a circle, adiameter that bisects a chord (that is not a diameter) is perpendicular to the chord. Theorem 12-8 In a circle, the perpendicular bisector of a chord contains the of the circle. A chord is a segment whose endpoints are on a circle. center P © Pearson Education, Inc., publishing as Pearson Prentice Hall. ©Pearson Education, Inc., publishing as Pearson Prentice Hall. Quick Check. Theorem 12-6 In a circle, a diameter that is perpendicular to a chord bisects the chord arcs and its . 1. If you are given that BC DF in the circles, what can you conclude? B 0 0 lO O lP; BC O DF O D P C F 䉺O 䉺P 2. Find thevalue of x in the circle at the right. 16 18 18 16 x 36 Q O 152 38 Geometry Lesson 12-2 Geometry Daily Notetaking Guide L1 L1 Daily Notetaking Guide Geometry Lesson 12-2 All-In-One Answers Version B 153 L1 Geometry: All-In-One Answers Version B (continued)

Name_____________________________________ Class____________________________ Date________________ Lesson 12-3 1 x 12 mDEF Topic: Relationships Among Geometric Figures Local Standards: ____________________________________ ( x 12 Vocabulary and Key Concepts. half themeasure of its intercepted arc. B 1 mAC 2 C ) The measure of an angle formed by a tangent and a chord is B half the measure of the intercepted arc. 1 1 2mBDC m⬔C B D Substitute. x° y 12 C C ( right congruent . angle. 3. The opposite angles of a quadrilateral inscribed in a circle are An inscribed anglehas a vertex that is on a supplementary A Intercepted arc circle and sides that are chords of the circle. . B C Inscribed angle intercepted arc is an arc with endpoints on the sides of an inscribed angle and its other © Pearson Education, Inc., publishing as Pearson Prentice Hall. Corollaries to the InscribedAngle Theorem © Pearson Education, Inc., publishing as Pearson Prentice Hall. y 1. Two inscribed angles that intercept the same arc are 0 m FG 70 G Inscribed Angle Theorem 0 ( y 12 m EF D 2. An angle inscribed in a semicircle is a Arc Addition Postulate D y° ) 70 1 y 12 mEFG Theorem 12-10 Allrights reserved. 80 75 Simplify. 90° 0 1 Because EFG is the intercepted arc of ⬔D, you need to find mFG in order 1 to find mEFG . The arc measure of a circle is 360, so 0 mFG 360 70 80 90 120 . All rights reserved. All rights reserved. A The measure of an inscribed angle is m⬔B ( C x Theorem 12-9:Inscribed Angle Theorem An 0 m EF F 80° Inscribed Angle Theorem 0 x 12 m DE 70° E Using the Inscribed Angle Theorem Find the values of x and y. NAEP 2005 Strand: Geometry Find the measure of an inscribed angle Find the measure of an angle formed by a tangent and a chord 2 Example.Inscribed Angles Lesson Objectives 1 Name_____________________________________ Class____________________________ Date ________________ 120 ) Arc Addition Postulate ) Substitute. 95 Simplify. Quick Check. In the Example, find m⬔DEF. 105 points in the interior of the angle. 154Geometry Lesson 12-3 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ © Pearson Education, Inc., publishing as Pearson Prentice Hall. Lesson Objectives 1 Find the measures of angles formed bychords, secants, and tangents 2 Find the lengths of segments associated with circles 0 Let mXY x. 1 Then mXAY NAEP 2005 Strand: Geometry Topic: Relationships Among Geometric Figures [( 12 ( 12 72 72 Key Concepts. intercepted arcs. m⬔1 All rights reserved. x° y° 1 1 (x 1 y) 2 (2) intersect outsidea circle is half the difference of the measures of the x° y° 1 y° 1 x° All rights reserved. x Theorem 12-11 The measure of an angle formed by two lines that A ) 2x) 360 ⫺ x 360 Theorem 12-11(2) x ] Simplify. 72 180 72 180 Add x to both sides. x 108 Solve for x. 108⬚ x Substitute. Distributive Property. arcwill be in the advertising agency’s photo. 2 Finding Segment Lengths A tram travels from 50 ft point A to point B along the arc of a circle with a radius of 125 ft. Find the shortest distance from point A to point B. y° 1 x° 155 x. 360 1 0 m⬔T 12 (mXAY mXY ) Local Standards:____________________________________ (1) intersect inside a circle is half the sum of the measures of the Geometry Lesson 12-3 Name_____________________________________ Class____________________________ Date ________________ Angle Measures and Segment Lengths Lesson12-4 Daily Notetaking Guide A x B x intercepted arcs. m⬔1 The perpendicular bisector of the chord AB contains 1 (x 2 y) 2 Theorem 12-12 For a given point and circle, the product of the lengths of the two segments from the point to the circle is constant along any line through the point and circle. a c b Pd abcd III. II. x w z y t P z (w x)w (y z)y (y z)y t 2 Examples. 1 Finding Arc Measures An advertising agency wants a frontal photo of a “flying saucer” ride at an amusement park. The photographer stands at the vertex of the angle formed by tangents to the “flying saucer.” What is the measure of the arc thatwill be in the photograph? In the diagram,0 the ) ) photographer 1stands at point T. TX and TY intercept minor arc XY and major arc XAY. 156 L1 Geometry Lesson 12-4 y P A 125 250 ft. The length of 2 the other segment along the diameter is 250 ft 50 xx 50 x2 x ft, or 200 ft. 200 10,000 100 Theorem 12-12(1) Multiply. Solve for x. The shortest distance from point A to point B is 2x or 200 ft. Quick Check. 1. Find the value of w. w° 250 2. Find the value of x to the nearest tenth. Y X 110° 70° 20 13.8 x 72° Daily Notetaking Guide All-In-One Answers Version B © Pearson Education, Inc., publishing asPearson Prentice Hall. © Pearson Education, Inc., publishing as Pearson Prentice Hall. diameter is I. 200 ft the center of the circle. Because the radius is 125 ft, the 14 16 T L1 L1 Daily Notetaking Guide Geometry Lesson 12-4 Geometry 157 39 Geometry: All-In-One Answers Version B (continued)Name_____________________________________ Class____________________________ Date________________ Lesson 12-5 Lesson Objectives 3 Graphing a Circle Given Its Equation Find the center and radius of the circle with equation (x 4) 2 (y 1) 2 25. Then graph the circle. (x 4)2 (y 1)2 25 (x4)2 (y 1)2 (5)2 NAEP 2005 Strand: Geometry Write an equation of a circle Find the center and radius of a circle Topic: Position and Direction Local Standards: ____________________________________ c c c h k r The center is Key Concepts. y Theorem 12-13 r (h, k) O x 2 8 6 4 2 O 6 8 x (x 8 0 )2 "5 ()2 y 2 Quick Check. Standard form )2 5 1. Write the standard equation of the circle with center (2, 1) and radius "2 . Substitute (ⴚ8, 0) for (h, k) and (x ⫹ 2) 2 ⫹ (y ⫹ 1) 2 ⫽ 2 Á 5 for r. Simplify. 2 Using the Center and a Point on a Circle Write the standard equation of a circle with center (5, 8) that passesthrough the point (15, 13). First find the radius. 2( h ) k x y )2 r #( Use the Distance Formula to find r. ⴚ15 5 8 #( )2 ( ⴚ13 )2 2 ) ⴚ20 ⴚ21 #( ( )2 Substitute (5, 8) for (h, k) and (ⴚ15, ⴚ13) for (x, y). Simplify. 441 #400 #841 29 Then find the standard equation of the circle with center (5, 8) and radius (x h )2 (y k)2 r 2 (x 5 )2 (y 8 )2 (x 5 ) 2 (y 8 ) 2 ( 29 841 29 © Pearson Education, Inc., publishing as Pearson Prentice Hall. )] 2 (y © Pearson Education, Inc., publishing as Pearson Prentice Hall. ⴚ8 (x ⫺ 2) 2 ⫹ (y ⫺ 3) 2 ⫽ 13 y 3. Find the center and radius of the circle with equation (x 2) 2 (y 3) 2 100. Then graph thecircle. 16 center: (2, 3); radius: 10 8 O . )2 (2, 3) 4 x ⴚ8 Standard form Substitute (5, 8) for (h, k) and 29 for r. Simplify. Geometry Lesson 12-5 Daily Notetaking Guide L1 L1 Name_____________________________________ Class____________________________ Date________________ Lesson 12-6Daily Notetaking Guide Geometry Lesson 12-5 159 Name_____________________________________ Class____________________________ Date ________________ 2 Describing a Locus in Space Describe the locus of points in space that are Locus: A Set of Points Lesson Objective 1 Draw anddescribe a locus 2. Write the standard equation of the circle with center (2, 3) that passes through the point (1, 1). All rights reserved. (x h )2 (y k )2 r 2 158 4 4 1 Writing the Equation of a Circle Write the standard equation of a circle with center (8, 0) and radius "5. ( 2 2 Examples. [x y 4 All rights reserved.(h, k) and radius r is (x h) 2 (y k) 2 r 2. and the radius is 5 . 6 All rights reserved. The standard form of an equation of a circle with center (⫺4, 1) (x, y) Relate the equation to the standard form (x ⴚ h)2 ⴙ (y ⴚ k)2 ⴝ r 2. 4 cm from a plane M. NAEP 2005 Strand: Geometry Topic: Dimension and Shape ©Pearson Education, Inc., publishing as Pearson Prentice Hall. 1 2 Name_____________________________________ Class____________________________ Date ________________ Circles in the Coordinate Plane Local Standards: ____________________________________ Vocabulary. 44 cm Alocus is a set of points, all of which meet a stated condition. M Examples. 1 Describing a Locus in a Plane Draw and describe the locus of points in a All rights reserved. All rights reserved. 44 cm 艎 Imagine plane M as a horizontal plane. Because distance is measured along a perpendicular segment froma point to a plane, the locus of points in plane that are 1.5 cm from 䉺 C with radius 1.5 cm. Use a compass to draw 䉺 C with radius 1.5 cm. space that are 4 cm from a plane M are a plane 4 cm above and parallel to plane M and another plane 4 cm below and parallel to plane M. Quick Check. * ) 1. Drawand describe the locus: In a plane, the points 2 cm from line XY. cm 5 1. cm The locus of points in the interior of 䉺 C that are 1.5 the center C. cm from 䉺 C is The locus of points exterior to 䉺 C that are 1.5 cm from 䉺 C is a circle with center C and radius 3 cm. © Pearson Education, Inc., publishing asPearson Prentice Hall. 5 1. C © Pearson Education, Inc., publishing as Pearson Prentice Hall. * ) 2 cm X 40 Geometry Lesson 12-6 Geometry Daily Notetaking Guide Y 2 cm 2. Draw and describe the locus of points in space that are equidistant from two parallel planes. A plane midway between the twoparallel planes, parallel to and equidistant from each. The locus of points in a plane that are 1.5 cm from a point on 䉺 C with radius 1.5 cm is point C and a circle with radius 3 cm and center C. 160 * ) Two lines parallel to XY , each 2 cm from XY . A L1 L1 Daily Notetaking Guide B Geometry Lesson 12-6All-In-One Answers Version B 161 L1 Data Analysis and Probability Workbook Answers Section 1 Graphs 8. Hours Spent Doing Homework 1. Number of 1 2 3 4 5 6 Tally | |||| ||| ||||| | | Frequency 1 4 3 6 1 1 Frequency page 1 Frequency Tables, Line Plots, and Histograms Rose Bushes 2. 1– 2–1.75 3–2.75 4–3.75 5–4.75 6–5.75 7–6.75 8–7.75 8. 75 Rose Bushes Wanted 8 7 6 5 4 3 2 1 ✗ ✗ ✗ Number of Hours ✗ page 3 Reteaching 1: Frequency Tables and Line Plots ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ 1 2 3 4 5 6 1. Inches 3 4 5 6 7 5 3 1 1 2 Frequency Charleston Rainfall ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ 3.Answers may vary. Sample: © Pearson Education, Inc. All rights reserved. Frequency Rose Bushes Wanted 3 9 8 7 6 5 4 3 2 1 0 2. Inches 5 6 7 0 1 2 3 4 5 2 0 4 1 Frequency San Francisco Rainfall ✗ ✗ ✗ ✗ ✗ ✗ ✗ 1–2 3–4 5–6 Number of Rose Bushes 0 page 2 Practice: Frequency Tables, Line Plots,and Histograms 1. 4 1 3. Inches Frequency ✗ ✗ ✗ ✗ ✗ 2 3 4 3 4 4 8 Boxes of Juice Sold ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ 26 27 ✗ ✗ ✗ ✗ 21 22 23 24 25 2. a student 3. 3 students 4. 13 students 5. 80 and 85, 75 and 90 6. No; the interval includes 2–2.75 h. 7. 10 students ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ 3 4page 4 Reteaching 2: Frequency Distributions 1. Interval 95–100 90–94 85–89 80–84 75–79 70–74 under 70 Data Analysis and Probability Teacher’s Guide Wilmington Rainfall Tally Frequency 2 4 2 7 4 4 2 Answers 11 Data Analysis and Probability Workbook Answers Interval Tally Frequency 60–79 80–99 100–119 120–139 140–159 160–179 180 and over 4 7 6 2 4 1 2 Tally 4 8 4 4 Thorpe 4. Daily Use of Petroleum in the U.S. 2.5 Frequency |||| |||| ||| |||| |||| 25 – 29 30 – 34 35 – 39 40 – 44 Ali Sports Figure page 5 Reteaching 3: Histograms 1. Answers may vary. Sample: Number of Raisins 30 25 20 1510 5 0 King 3 4 3 5 4 2 3 Robinson 2.0 1.5 1 0.5 0 Year 5. Answers may vary. Sample: Daily petroleum use has been declining since 1970. 8 7 6 5 4 3 2 1 0 page 8 Stacked Bar and Multiple Line Graphs 1. grade 7 2. grade 7; grade 8 3. Sample: Find the total, subtract the bottom part. 4. 1993 5. Find thenumber of each item sold and add the two. page 9 Practice: Stacked Bar and Multiple Line Graphs 44 40 – 34 39 35 – 25 – 29 1. 30 – Frequency 2. Average Viewing Time 8:00 P.M.–11:00 P.M. (Mon. – Sun.) Males Females 55 Number of Raisins page 6 Bar and Line Graphs 1. tens 2. fives 3. thousands4. hundreds 5. fives 6. tens 7. Bar graph; data shows amounts, but not changes over time. 8. Line graph; data shows change over time. page 7 Practice: Bar and Line Graphs 1. names of the athletes 2. Answers may vary. Sample: 5 12 Answers 25 – 54 18 – 24 14 12 10 8 6 4 2 0 2 4 6 8 10 12 14 Time(Hours) Data Analysis and Probability Teacher’s Guide © Pearson Education, Inc. All rights reserved. 3. All-Time Favorite Sports Figures Zaharias 700–749 750–799 800–849 850–899 900–949 950–999 1000 and over 3. Ruth Frequency 196 0 196 5 197 0 197 1985 1980 5 199 1990 2005 0 Tally Numberof Votes Interval Millions of Barrels 2. ;; ; ; ;; ;; ; ;;;; ;; ; ;; Data Analysis and Probability Workbook Answers Average Viewing Time 8:00 P.M. – 11:00 P.M. (Mon. – Sun.) Time (Hours) 2. 15 12 9 6 3 0 18 – 24 25 – 54 Ages Male Misc. Gas Food 10. Children per Family Population (100,000s) Population 5.04.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 6. Population Population (100,000s) Clothes Rent Female 1960 1970 1980 1990 2000 Year Hialeah, FL Birmingham, AL © Pearson Education, Inc. All rights reserved. Budget Phone 55 3. double bar graph 4. sliding bar graph 5. 9. 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.51960 1970 1980 1990 2000 Year Hialeah, FL Birmingham, AL 1 2 0 5 4 3 page 11 Practice: Circle Graphs 1.–4. Sample graph shown. Category Amount Budgeted Percent of Degrees in Total Central Angle 1. Gas $200 10% 36° 2. Meals $400 20% 72° 3. Motels $500 30% 108° 4. Other $800 40% 144°Rahman Family Budget Gas 10% Meals 20% Other 40% Motels 30% 7. multiple line graph page 10 Circle Graphs 1. 72° 2. 144° 3. 180° 4. 18° 5. 126° 6. 36° 7. 72° 8. 30° Data Analysis and Probability Teacher’s Guide Answers 13 Data Analysis and Probability Workbook Answers 5–8. Sample graphshown. Category Amount Budgeted Percent of Total Degrees in Central Angle 5. Clothing $50 40% 36° 6. Entertainment $40 32% 72° 7. Savings $25 20% 108° 8. Transportation $10 8% 144° Transportation 8% Savings 20% Milk Drunk not sure less 13.1% 10.1% more 56.3% the same 20.5% page 12Reteaching: Circle Graphs 1. 97° 2. 86° 3. 47° 4. 65° 5. 65° page 13 Stem-and-Leaf Plots 1. Times for the 200-M Dash 38 37 36 35 34 33 32 31 30 29 68 569 03447 568 3 8 means 38 56 3 0479 23 4777 4 1225 479 2 35 7 01459 8 1226 9 0169 9 9 means 9.9 5. 8.1; 8.2 and 6.3; 3.6 6. 4 17 36 70 5 6 7 821 26 86 34 75 92 19 17 4 17 means 417 6. 586; no mode; 400 7. 17 4 6 9 18 5 5 6 19 4 5 20 4 17 4 means 17.4 7. 18.5; 18.5; 3 8. 15° 9. 64° 10. 69° 11. 76; 75; 79; 80 page 15 Reteaching: Stem-and-Leaf Plots 1. 1 5 6 2. 64; 15; 49 2 3 4 5 6 47 669 25 149 134 1 5 means 15 3. 14 7 8 9 8 234456679 9012 7 8 means 78 38 8 means 38.8 4. 45 4 5 6 7 8 013 5678 235666 114 1235 8 1 means 81 2. 34.7 s 3. 9.3 s 4. 34.7 s 5. 16 students 14 Answers Data Analysis and Probability Teacher’s Guide © Pearson Education, Inc. All rights reserved. Entertainment 32% 9. 4. 0 1 2 3 4. 23; 24; 32 5. 6 3 3 Lucy’sBudget Clothing 40% page 14 Practice: Stem-and-Leaf Plots 1. 10, 11, 12, 13 2. 4, 5, 9, 9 3. 115; 104 and 115; 38 Data Analysis and Probability Workbook Answers page 16 Activity: Relating Stem-and-Leaf Plots to Histograms 16 a. Stem 3 4 5 6 7 b. 27 4467 78 1348 69 5 10 15 7. 60 65 70 75 80 85 908. 0 Grouping Intervals Frequency 30–39 40–49 50–59 60–69 70–79 2 4 2 4 2 1st 5 10 15 20 25 30 35 40 45 50 Set 2nd Set page 20 Reteaching: Box-and-Whisker Plots 1. 1 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 5 5 6 7 7 7 8 8 8 9 9 10 11 13 2. median 4 3. lower median 3; upper median 8 4. 4 1 2 3 4 1 3 4 5 67 8 9 10 11 12 13 8 13 3 2 1 0 30–39 40–49 50–59 60–69 70–79 Score page 17 Puzzle: Keeping Score © Pearson Education, Inc. All rights reserved. 0 Leaf Frequency Table Number of Teams c. 6. Stem 5 6 7 8 7 Leaf .75, .90 .35, .45, .75 .20, .50, .50, .40, .75 .75, .75, .75 .20 means 7.20 90 110 130150 170 2. 13, 4, 21 0 10 20 30 40 page 19 Practice: Box-and-Whisker Plots 1. 55 miles, 15 miles 2. 35 miles 3. 75% 4. 6 runners 5. 10 15 20 page 22 Practice: Reading and Understanding Graphs 1. oxygen 2. oxygen, carbon, hydrogen 3. There are other elements in quantities too small to be labeledindividually. 4. Western Europe 5. South and Central America, Middle East, Africa 6. North America 7. A greater percentage of men are reaching ages 25–29 without having married. 8. 1960 to 1970 9. C page 23 Reading Graphs Critically 1. the first graph 2. about 3 times 3. The second graph; since thescale is smaller, the bars can be read more accurately. 4. By using the break, most of the bar for the United States has been left out. page 18 Box-and-Whisker Plots 1. 98, 80.5, 118 70 page 21 Reading and Understanding Graphs 1. 20 books 2. January 3. 10 books 4. October and November 5. Englishand Spanish 6. 12 people 7. 12 people 8. Polish 25 30 Data Analysis and Probability Teacher’s Guide page 24 Practice: Reading Graphs Critically 1. 45–64 year olds 2. 45–64 year olds 3. Answers may vary. Check students’ graphs. 4. Answers may vary. 5. Answers may vary. Check students’ graphs. 6.Answers may vary. Check students’ graphs. 7. Answers may vary. 8. Answers may vary. Sample: to convince people that one program is more popular than it is page 25 Activity: Organizing and Analyzing Data 1. no; too wide a range of data 2. D.C., N.J., and R.I. 3. possible answer: 100 people persquare mile 4. People might assume that the two places have similar densities. 5. You could list the three places individually. 6. nothing Answers 15 Data Analysis and Probability Workbook Answers page 26 Scatter Plots and Trends page 29 Activity 2: Making a Scatter Plot 1. 1. Test Scores and TVLatitude (°N) 100 90 80 70 50 30 20 2. Negative; as one value goes up, the other goes down. 3. The more TV students watch, the lower their test scores. page 27 Practice: Scatter Plots and Trends 1. positive trend 2. negative trend 3. no trend 4. Students’ graphs should show a negative trend. Sample:Residents of Maintown 1,200 1,150 1,100 1,050 1,000 950 900 70 50 40 30 20 7a. ERA 5.00 4.00 3.00 60 70 80 90 100 Wins b. negatively correlated page 31 Reteaching: Scatter Plots and Trends 1. 6. Answers may vary. Sample: time spent skiing down a hill vs. speed of skier page 28 Activity 1: Makinga Scatter Plot 1–2. Check student’s work. 3–4. Check students’ work. There should be a positive correlation. Answers Height in Inches 5 10 15 20 25 30 35 40 45 50 Time Spent Studying (minutes) Weight and Height Survey 70 60 50 40 30 30 40 50 60 70 80 90 100 110 Weight in Pounds 2. about 61 in.3. about 63 lb 4. yes 5. Sample: As height increases, weight increases. Data Analysis and Probability Teacher’s Guide © Pearson Education, Inc. All rights reserved. Studying Time vs. Number of Correct Words 22 20 18 16 14 12 10 8 6 4 2 0 2.0 3.0 4.0 5.0 6.0 Precipitation (in.) page 30 Analyzing ScatterPlots 1. negatively correlated 2. unrelated 3. positively correlated 4. positively correlated 5. negatively correlated 6. unrelated 5. Students’ graphs should show a positive trend. Sample: Number of Words Spelled Correctly 40 50 60 Temp (°F.) 4. no 1,800 1,900 2,000 2,100 Homeowners 16 30 3. 0 1 2 3 45 Hours of TV Vacation Homeowners 40 60 Latitude (°N) Test Scores 2. yes, no 50 Data Analysis and Probability Workbook Answers page 32 Assessment 1 1. 34 2. biking; 12 students 3. sailing and hiking; 4 students 4. 24% 5. Because students picked activities besides the ones listed, but too fewstudents picked those activities for each activity to get a separate bar. 6. 100, 50, no mode 7. 0 20 0 20 40 60 50 80 100 80 100 Section 2 Measures of Central Tendency page 33 Measures of Central Tendency 1. mean: 3; median: 3; mode: 4 2. mean: 8.1; median: 8; mode 5 3. mean: 5.25; median: 5;mode 0, 1 4. mean: $3.25; median: $3.25; mode: $4 5. Median; there is no mode and the median is greater than the mean. 6. Mode; both the median and the mean are less than the mode. page 34 Practice 1: Measures of Central Tendency 1. 1.5; 2; 2 2. 3.3; 1; 0 3. 29.1; 29.5; 25, 30, and 35 4. 8.1; 9; 9 5.73.6; 72; no mode 6. 153.2; 146; no mode 7. Mode; data are nonnumerical. 8. Mode; data are nonnumerical. 9. Mean; there should be no outliers. 10. Mean or median; use median if there are outliers. 11. 92 12. at least an 87 13. 85 14. 86 © Pearson Education, Inc. All rights reserved. page 35 Practice 2:Mean, Median, Mode, and Outliers 1. 23 students 2. at least 12 students; up to 11 students 3. mean: 287.5, median: 300, mode: 200 4. 50 5. lower 6. The median, because there is an outlier. 7. 8 points 8. 23 points 9. Answers will vary. page 36 Activity 1: Choosing an Appropriate Measure 1. Checkstudents’ work. 2. a, b, c, d, g 3. all of them, where the mode exists 4–6. Check students’ work. page 37 Activity 2: Average Temperature 1. Stem Leaf 5 8 6 0233 7 0444555667778888899 8 000011223455666777888899999 9 016 10 page 38 Activity 3: Wink Count 1–6. Answers depend upon class data.EXTRA 7. They must score at least 12 runs during the 3 games. One way to find the different ways in which this can be done is to make a chart. Game 1 Game 2 Game 3 12 0 0 0 12 0 0 0 12 11 1 0 11 0 1 1 11 0 etc. There are 91 ways in all in which the additional 12 runs may be scored. page 39 Puzzle:Mean, Median, and Mode 1. {2, 4, 5, 7, 7} 2. All should have found the same set. 3. One 4. a. 0 b. 1 c. 0 d. 0 e. 0 5. Several solutions are possible, including {5, 5, 5, 5, 5}, {4, 5, 5, 5, 6}, {3, 5, 5, 5, 7}, etc. 6. Conditions: 0 a b c d e 20, and 2a d e 4c. Several solutions are possible, including {0, 0, 0, 20, 20}.7. Conditions: a b c d e 5c, and a b c d e mode. One possible solution is {0, 0, 5, 5, 5}. 8. no solution 9. Answers may vary. Samples: {0, 0, 0, 20, 20} and {0, 0, 20, 20, 20} page 40 Reteaching: Mean, Median, and Mode 1. 12, 11.5, none 2. 87¢, 86.5¢, none 3. 231, 231, none 4. 50, 49, 43 5. $130, $129,none page 41 Assessment 2 1. 3:03 2. 2:04; 1:55 3. 1:58 4. 2:03 5. Answers may vary. Sample: The median, because it eliminates the effect of the outlier and gives a representative time. 6. Answers may vary. Sample: The mode, because it is arbitrary that two skiers happened to have the exact time. 7.1:18; the difference between the fastest and the slowest time. 8. below; 2.3 degrees 9. 41 degrees 10. 63 degrees Section 3 Use and Misuse of Data Displays page 42 Choosing an Appropriate Graph 1. multiple line graph: shows changes in two sets of data over time 2. circle graph: shows how the club’sbudget is divided into parts 3. double bar graph: compares two sets of data 4. circle graph: shows how 100% is divided into parts 5. line graph: shows change over time 27 2. Answers will vary. 3. mean 81.05; median 80.5; mode 78 and 89 (bimodal) No city has the mean, although Columbus and Seattleare close. No city has the median, although Albany, Pittsburgh, St. Louis, and Salt Lake City lie just below, and Columbus and Seattle lie just above. The data is bimodal: Denver, Detroit, Spokane, Topeka, and Tulsa all have 78; Albuquerque, Portland, Raleigh, Richmond, and San Antonio all have 89. 4–6. Answers depend upon class data. Data Analysis and Probability Teacher’s Guide Answers 17 Data Analysis and Probability Workbook Answers 7 6 5 4 3 2 1 1–5 als irds tiles bians Fish nails eans m B ep hi S tac m us Cr p Am Type page 44 Activity: Appropriate Graphs 1. line graph 2. bar graph 3. no4. Bar graph; it compares amounts. page 45 Misleading Graphs 1. graph A 2. graph B 3. The years on the horizontal axis are spaced differently; graph A seems to rise more steeply than graph B. 4. graph D 5. graph C 6. vertical scale is unbroken on one, and broken on the other; difference in the barsseems greater on graph D 7 6 5 4 3 2 1 1–4 10 15–24 25–30 page 48 Using Graphs to Persuade 1 1. no 2. 2.0 to 120 3. Answers may vary. Sample: • • Number of Clear Days 14 13 11–14 Years Average Number of Clear Days per Year 2. • 5–10 2. Lois’ graph has intervals of 5 years and Harold’s hasintervals of varying numbers of years. Lois’ is more useful. 3. You may wish to have students compare their graphs. page 46 Practice: Misleading Graphs 1. 15 11–15 16–20 21–25 Years Harold’s Graph Teachers’ Years of Service 150 100 50 0 J F M J F M 3. 2 cars 4. Probably not; car sales vary greatlyand Cities 4. Answers may vary. Sample: Average Number of Clear Days per Year Number of Clear Days information from the first three months is not necessarily indicative of the whole year. 5. The vertical axis changes units; the graph continues through a break. 6. The number of people who preferYummy Cereal is increasing sharply. Albany Atlanta Boise Houston St. Louis Seattle R 6–10 125 100 75 50 25 0 Albany Atlanta Boise Houston St. Louis Seattle a Lois’s Graph Teachers’ Years of Service Number of Teachers Number 80 70 60 50 40 30 20 10 M 1. Number of Teachers U.S. EndangeredAnimals page 47 Activity: Analyzing Graphs Cities 18 Answers Data Analysis and Probability Teacher’s Guide © Pearson Education, Inc. All rights reserved. page 43 Practice: Choosing an Appropriate Graph 1. private owners 2. about 25,000 people 3. line graph, shows change over time 4. bar graph,compares quantities 5. scatter plot, shows a relationship between sets of data 6. circle graph, compares parts of a whole 7. bar graph: See students’ graphs. Sample: Data Analysis and Probability Workbook Answers 6. U.S. Union Membership 20 18 16 14 12 10 8 6 4 2 0 page 49 Using Graphs toPersuade 2 Union members (millions) 1.60 1.50 1.40 1.30 1.20 1.10 199 6 199 7 199 8 199 9 200 0 200 1 Year 7. page 50 Practice: Using Graphs to Persuade 1. birds 2. no 3. the break in the vertical axis. U.S. Endangered Species Number of Species © Pearson Education, Inc. All rights reserved. 4. 8070 60 50 40 30 20 10 Mammals Birds Type Fish 5. The differences seem much less. Data Analysis and Probability Teacher’s Guide 90 19 80 19 70 19 60 40 19 30 3. The first graph implies prices decreased rapidly from 1997 to 1998, then increased rapidly from 1999 to 2000. The second graph impliesslower changes. 20 18 16 14 12 10 8 6 4 2 0 19 Year U.S. Union Membership Union members (millions) 1 200 0 200 199 9 8 199 199 199 7 1.60 1.50 1.40 1.30 1.20 1.10 6 Price (dollars) Gasoline Retail Prices 19 2. 19 3 19 0 4 19 0 50 19 6 19 0 70 19 8 19 0 90 Year 50 Price (dollars) Gasoline RetailPrices 19 1. Year 8. The horizontal scales are different. page 51 Analyzing Games and Making Predictions 1. No; it suggests that Player A wins 3 and Player B wins 5. 2. HTHH, THTT, HTTH, HTHT, HHHT, THHT, TTHT, TTHT, HTHT, TTTH; Player A wins 4 and B wins 6. 3. No: Player A wins 7 andPlayer B wins 13. 4. Yes; in 16 rounds, both players are likely to win 8 points. page 52 Practice: Analyzing Games and Making Predictions 1. not fair 2. fair 3. fair 4. not fair 5. Possible answers: 0–1 for correct, 2–9 for incorrect 6. Check students’ work. 11 50 22% if answer shown here for Exercise 5 isused. 7. possible answer: 0–4 for correct, 5–9 for incorrect 29 8. Check students’ work. 50 58% if answer shown here for Exercise 7 is used. 9. Possible answer: 0–7 for correct, 8–9 for incorrect 10. Check students’ work. 41 50 82% if answer shown here for Exercise 9 is used. Answers 19 Data Analysisand Probability Workbook Answers 2. 1st choice 2nd buttons small snaps buttons medium snaps Lawns Mowed 0 1 2 3 4 Week Number 2. Answers will vary. Sample: 5 4 3 2 1 0 buttons large snaps 0 1 2 3 Week Number Section 4 Counting Principles page 54 Counting Outcomes 1. 12 2. 15 3. 16 4. 205. 12 6. 30 page 55 Practice: Counting Outcomes 3rd choice choice milk pizza juice milk spaghetti juice blue beige blue beige blue beige blue beige blue beige blue beige 3. 45,697,600 license plates 4. 6,760,000 license plates 5. 24 types 6. 19,656 results 7. 1,000 codes 8. 358,800 passwords 4 3.Answers will vary. Graph A is good for giving the impression that something is decreasing rapidly. Sample: showing the change in a bank account over time to explain why a bigger allowance is needed. 4. Answers will vary. Graph B is good for giving the impression that something is decreasing moreslowly. Sample: showing the change in a bank account over time to convince one’s self that a job is needed. 1. 1st choice 2nd 3rd choice choice 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 pudding apple pudding apple pudding apple pudding apple page 56 Permutations 1. 24 2. 6 3. 4 4. 90 5. 1 6. 240 7. 120 8. 20 9.1,320 10. 24 11. 210 12. 360 13. 6P2 30 14. 10P3 720 page 57 Practice: Permutations 1. 720 2. 479,001,600 3. 362,880 4. 336 5. 79,833,600 6. 15,120 7. 56 8. 1,814,400 9. 120 10. 3,603,600 11. 24 ways 12. 5,040 ways 13. 40,320 orders 14. 362,880 arrangements 15. 68,880 ways 16. 2,730 ways 17.1,406 ways 18. 151,200 arrangements 19. 17,576 arrangements 20. 3,628,800 arrangements page 58 Combinations 1. 20 2. 10 3. 21 4. 4 5. 28 6. 15 7. 126 8. 10 9. 6 10. 35 11. 56 12. 84 13. 10C3 120 14. 6C2 15 15. 7C2 21 16. 15C5 3,003 17. 10C5 252 18. 8C3 56 page 59 Practice 1: Combinations 1.9 2. 70 3. 330 4. 330 5. 1 6. 84 7. 924 8. 28 9. 120 10. 6 11. 70 12. 21 13. 210 ways 14. 490,314 ways 15. 5,005 ways 16. 1,370,754 ways 17. 66 handshakes 18. 10,626 committees 19. B, J; B, E; B, M; B, X; J, E; J, M; J, X; E, M; E, X; M, X 20. D page 60 Practice 2: Permutations and Combinations 1. 422. 21 3. 336 4. 3,024 5. 3 6. 210 7. a. 24 b. 120 c. 24 d. 51 8. 10 9. 42,840 10. 30 11. 120 12. 360 13. 720 14. 720 page 61 Assessment 4 1. 5040 2. There would be 8 times as many (or 40,320). 3. 15 4. 15 or 20% 5. 60 ways 6. Answers will vary. Sample: 6P3 could represent how many ways you couldmake a 3-digit combination out of the numbers 1 to 6; 6C3 could represent how many ways you select three students from a group of six. 7. 24C6, 24P6, and 24! 20 Answers Data Analysis and Probability Teacher’s Guide © Pearson Education, Inc. All rights reserved. Lawns Mowed page 53 Assessment3 1. Answers will vary. Sample: Data Analysis and Probability Workbook Answers page 67 Practice: Sample Spaces Section 5 Theoretical Probability page 62 Theoretical Probability 3 1. 1; 0.2; 20% 2. 5; 0.4; 40% 3. 12, 0.5, 50% 4. 10 , 0.3, 30% 9 7 10 8 3 22 2 5. 10 , 0.9, 90% 6. 11 7. 11 8. 11 9. 25 10.25 11. 25 page 63 Practice: Theoretical Probability 1 1. 10 ; 0.1; 10% 2. 12; 0.5; 50% 3. 1; 1.0; 100% 4. 15; 0.2; 20% 6 8 5. 23 or 34.8% 6. 23 or 26.1% 7. Add a marble that is not blue. 8. 21 9. 14 10. 41 11. 43 12. 43 13. 21 14. 43 15. 34 997 3 16. 12 17. 21 18. 34 19. 43 20. 1000 , 0.3%, 0.003 21. 1000 ,99.7%, 0.997 page 64 Reteaching: Theoretical Probability 3 1 1 1 1 1. 51 2. 52 3. 51 4. 10 5. 10 6. 0 7. 10 8. 12 9. 12 1 2 3 4 5 6 7 8 9 10 A A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 B B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 C C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 D D1 D2 D3 D4 D5 D6 D7 D8 D9 D10 E E1 E2 E3E4 E5 E6 E7 E8 E9 E10 F F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 G G1 G2 G3 G4 G5 G6 G7 G8 G9 G10 H H1 H2 H3 H4 H5 H6 H7 H8 H9 H10 2a. 10. 61 11. 23 12. 16 13. 0 14. 13 H page 65 Activity: Finding Errors in Statistical Analysis 1. 2 Incorrect probability for each event P(even) 5 P(2) 1 P(4) 1 P(6) T 55 1 1 1 6 1 6 1 6 3 1 6 , or 2 2 5 5 1 8 2 10 2 10 6 3 10 , or 5 H T H T H T H T H T 1 21 1 21 2b. 14 3. 12 kinds 4. 32 5. 64 6. 64 7. 64 page 68 Counting Outcomes and Probability 1. 18 possible outcomes 2. C1-S1-F1, C1-S1-F2, C1-S1-F3, C1-S2-F1, C1-S2-F2, C1-S2-F3, C1-S3-F1, C1-S3-F2, C1-S3-F3, C2-S1-F1, C2-S1-F2, C2-S1-F3, C2-S2-F1, C2-S2-F2, C2-S2-F3, C2-S3-F1, C2-S3-F2, C2-S3-F3 3. 31 4. 91 P(,4) 1 P(odd) 2 P(, 4>odd) 5 (P(1) 1 P(2) 1 P(3) 1 P(1) 1 P(3) 1 P(5) 1 P(7) 1 P(9)) 2 P(1) 1 P(3)) 1 1 1 1 1 1 1 1 5 Q10 1 10 1 10 1 10 1 10 1 10 1 10 1 10 R 1 Q10 H T 2. 1 Cannot countthe same event twice © Pearson Education, Inc. All rights reserved. 1. page 69 Practice: Counting Outcomes and Probability 1. m1A, m1B, m1C, m2A, m2B, m2C, m3A, m3B, m3C, m4A, 1 m4B, m4C 2. 31 3. 12 4. 6 choices: AM, AN, BM, BN, CM, CN 5. 8 combinations; P1C1; P1C2; P2C1; P2C2; P3C1;P3C1; P3C2; P4C1, P4C2 6. 140 routes 7. 468 combinations. 1 10 R median is the middle number once the data set is in order. 9 Mean: 64 11 , or 5 11 Median: 6 Mode: 7 page 70 Independent and Dependent Events 9 3 6 4 1 1 1. 64 32 2. 64 3. 64 16 4. 91 5. 61 6. 12 7. independent 8. independent 9.independent 10. dependent page 66 Sample Spaces 1. 12 page 71 Practice 1: Independent and Dependent Events 6 1 1 1 1. 51 2. 15 3. 10 4. 0 5. 41 6. 81 7. 41 8. 32 9. 11 3. Labels on the horizontal and vertical axes are transposed 4. 1 2 3 Terms are not written in numerical order; the 1 2 H T 2. 10 H 3T H T m a e i 4 H 5 T H T H a e i T second guest’s choice is limited by the first guest’s choice. 15. Independent; the second flip is not affected by the first. 1 1 16. 81 17. 72 n o u 5 1 1 1 10. 22 11. 11 12. 33 13. 11 14. Dependent; the 6 page 72 Practice 2: Independent and Dependent Events o u 3. 32 4.54 5. 9 6. 24 Data Analysis and Probability Teacher’s Guide 9 6 6 6 4 2 4 1 1. 25 2. 25 3. 25 4. 25 5. 55 6. 55 7. 55 8. 221 9. 61 2 10a. 65 10b. 42 65 Answers 21 Data Analysis and Probability Workbook Answers trash > > /¶2> > /¡``/3``£/ 1¡/```/`£0/ brown /¡``4/``£/ 5/¶ /¡``/9``/£ black laundry brown blackbrown black 6 43 17 11b. 91 12. 100 ; 200 page 73 Reteaching: Independent and Dependent Events 9 3 3 1 1 1 1 1 1. 25 2. 50 3. 100 4. 100 5. 45 6. 15 7. 30 8. 10 page 74 Analyzing Events Not Equally Likely 7 3 2 1 1. 12, 12, 20 , 4, 5, 4 2. 1 3. 8 4. 6 page 75 Practice: Analyzing Events Not EquallyLikely 1 1 1 1 1 1 1 1. 13 2. 13 3. 13 4. 13 5. 13 6. 13 7. 26 8. 41 9. 1 2 16. 10. 3 13 11. 1 52 12. 3 4 13. 51 52 14. 2 7 15. 8 19 7 16 page 76 Activity: Analyzing Events Not Equally Likely 9 19 2 1. 47 2. 0.77 3. 0.9999 4. 52 5. 13 6. 13 page 77 Puzzle: Analyzing Theoretical Probability 1. 53 2. 25 3. You aremore likely to pick a 3. 4. The incorrect assumption is that there is a 13 chance of getting a 5 with two number cubes. With two number cubes there are 36 possible outcomes. Of those, 11 contain at least one 5, so your 11 chances are 36 , which is less than 31. With 6 number cubes, chances are 1Q56R< 67%. 6 page 82 Activity 2: Theoretical and Experimental Probability 1–2. Check students’ work. 3. no 4. 16 5. no 6. For a large number of rolls, you will likely roll a 3 one-sixth of the time. 7. Since P (any number from 1 through 6) 1, students should be able to predict that this result will occur 12 times.Since P(7) 0, students should be able to predict that this result will occur 0 times. 8. Yes, because through repeated trials the experimental results should approach the theoretical probability. 9–10. Check students’ work. page 83 Activity 3: Increasing the Number of Trials Check students’ tables. 1. 40 2.Answer should be close to 24. 3. answer should be about 3.16. 4. Check students’ calculations. page 84 Predictions Based on Experimental Probabilities 8 24 1. P(broken) 600 0.04 2. Yes; P(broken) 500 0.016 3. 0.04 18,000 720 crayons 4. 0.016 18,000 288 crayons 5. 98.4% page 85 Practice:Predictions Based on Experimental Probabilities 3 4 1. 200 2. yes; P(defective) now is 375 , which is less than 3 . 3. 175 g 4. 250 g, 100 g 5. 150 g 6. The mass must 200 be within 75 g of 175 g. 7. 87 cars 8. 8,000 bearings page 86 Reteaching: Predictions Based on Experimental Probabilities 9 x x 8 1.160 shirts, 400 2. 144 shirts, 500 5 8000 5 8,000 16 3. 480 games, 400 5 x 12,000 x 12,000 30 4. 450 games, 800 5 x 12,000 65 x games, 1,700 5 12,000 page 78 Assessment 5 1 1. They are equally likely: 21 3 12 3 12 3 12 5 16 . 2. There are 6 ways of getting two heads and two tails: HHTT; HTTH;HTHT; THHT; THTH; TTHH, but only one way of getting 1 HEADS TAILS HEADS TAILS. 3. 21 or 50% 4. 270 , 725 19 1 1 5. 28, 561 6. 27 or 70% 7. 3 8. The weather for two days in a row is not an independent event. If we get the storm it may be more likely to rain both days. If the storm misses us, itprobably won’t rain either day. 19 5. 456 games, 500 5 Section 6 Experimental Probability page 88 Random Samples and Biased Questions 1. C 2. B 3. No; you are more likely to interview homeowners. 4. No; you are more likely to interview renters. 5. Yes; you can’t tell if people own or rent. page 79Theoretical and Experimental Probability 9 1 3 1 3 1 4 1 4 2 2 1. 25 ; 5 2. 25 ; 15 3. 25 ; 3 4. 17 75; 15 5. 15 ; 15 6. 10 7. 2 3 1 8. 20 9. 20 page 80 Practice: Theoretical and Experimental Probability 9 5 5 7 1. 31 2. 12 3. 41 4. 32 5. 12 6. 43 7. 14 8. 14 9. 92 8 4 10. 15 11. 31 12. 45 13. 4 to 11 14. 8 to 3715. 32 1 1 16. 79 17a. 1200 ; 1 to 1199 17b. 400 ; 1 to 399 1 18. 0.55; 90 slices 19. 10,000 ; 1 to 9,999 page 81 Activity 1: Theoretical and Experimental Probability 1a-b. Check students’ work. 1c. Answers may vary; graphs should approach 50% as number of tosses increases. 22 Answers 6. 459 page 87Assessment 6 9 3 1 3 1 1 1. 100 , 1 to 99 2. 38, 52 3. 14, 40 4. 10 , 4 5. 40 ,8 x 1 1 6. 10 or 10% 7. 100,000 disks 8. 10 disks 9. 25 ; 5 500 20 students Section 7 Statistical Investigations and Simulations page 89 Practice: Random Samples and Biased Questions 1.–10. Answers may vary. Samples aregiven. 1. random sample 2. Not a random sample; students who use the vending machine may not represent all types of students. 3. fair 4. Biased; Do you prefer hardwood floors in your home? 5. fair 6. Biased; How many servings of fruits and vegetables do you eat? 7. Biased; Do you prefer thickcarpeting? 8. fair 9. fair 10. Biased; Does TV news portray life accurately? Data Analysis and Probability Teacher’s Guide © Pearson Education, Inc. All rights reserved. 11a. Data Analysis and Probability Workbook Answers page 90 Planning A Survey 1. No; the sample includes only students interested inart, not the whole population. 2. Yes; every student has an equal chance of being selected. 3. Closed-option; Which sport do you think is the most exciting? 4. Open-option; Did you enjoy the book we just read in English? 5. Questioner is suggesting a preferred answer. 6. Question assumes that the moviewas boring. page 91 Practice 1: Planning a Survey 1. shoppers in a mall who are at least 16 years old 2. 966 never, 644 occasionally, 536 regularly 3. 2,146 people 4. no 5. open 6. The question makes gourmet meals seem better than plain ones. 7. The question makes shortened school days seeminferior. 8. No. People who work full-time are more likely to shop after work than midday. 9. Answers may vary. Some things to consider are using good questions, getting a large enough random sample, and avoiding biased questions. page 92 Practice 2: Random Samples and Surveys 1. 320 students 2.352 students 3. 200 students 4. 192 students 5. The views of people coming out of a computer store may not represent the views of other voters. This is not a good sample because it is not random. 6. The city telephone book may cover more than one school district. It would also include people who donot vote. This is not a good sample because it does not represent the population. 7. This is a good sample. It is selected at random from the population you want to study. © Pearson Education, Inc. All rights reserved. page 93 Conducting a Statistical Investigation 1. Check students’ work. page 94Investigation 1: Happy Birthday 1–6. Check students’ work. EXTRA: James Polk (1796) and Warren Harding (1865) were born on November 2. John Adams (1826), Thomas Jefferson (1826), and James Monroe (1831) all died on July 4. page 95 Investigation 2: Hits and Misses (Geometric Probability) 2 1.A 5 pr2 2. 2r ? 2r 5 4r2 3. pr2 5 6–8. Answers may vary. 4r p 4 4. 0.785 page 96 Investigation 3: Sticky Dot Number Cubes 1 2 1–5. Answers may vary. 6. P(0) 5 36 , P(1) 5 36 , 3 3 5 4 P(2) 5 36 , P(3) 5 36 , P(4) 5 36 , P(5) 5 36 , 5 3 3 4 P(6) 5 36, P(7) 5 36, P(8) 5 36, P(9) 5 36, 2 1 P(10) 5 36 , P(11) 536 , page 97 Planning a Simulation 1. the probability that out of 3 children, one will be a girl 2. a coin toss 3. three tosses of a coin 4. the probability that out of 6 tees, there will be 2 red, 2 white, and 2 blue 5. a number cube with 2 numbers assigned to each color 6. 6 rolls of the number cube 7. a numbercube with 2 numbers assigned to each color 8. 12 rolls of the number cube Data Analysis and Probability Teacher’s Guide page 98 Simulation 1: A Multiple Choice Test 1. 15 2. Answers may vary. Theoretically, 8.4 rolls 3. Answers 51 may vary. The theoretical probability is 243 . 5. Answers may vary. Thetheoretical value is 20. 6. Answers may vary. The theoretical probability is 21. 7. Answers may vary. The theoretical value is 7.5. 8. Answers may vary. The theoretical 3 probability is 16 . 9. Probability of getting 4 or more answers correct. page 99 Siumulation 2: Dyeing T-Shirts 1. 12 2. 52 3. 18 4. 35 5.Answers may vary. Theoretically, 11.25 shirts. 6. Answers may vary. The theoretical probability is 38. 7. Answers may vary. The theoretical value is 18.75. 8. Answers may vary. The theoretical probability is 58. 9. twice. Answers may vary. Theoretically, 1.875 times. 1 1 10. 15 . Answers may vary. Thetheoretical probability is 16 . page 100 Simulation 3: A Fair Game? In game 1, there is 1 chance out of 3 to match. In game 2, chances are 1 out of 2 that there will be a match. page 101 Analyzing Simulations 5 3 3 1 4 1. m n 2. 5 3. 5 4. 1 5. 15 6. 8 7. 25, 15 8. 8 , 8 3 9. 40 10. 12 11. No. The probability ofeach person winning the lottery is much less than 12. page 102 Reteaching 1: Simulations 1. Sample: Use a set of number cards 1–10 and draw one card at a time from a bag. Then replace the card. 2. Sample: Roll a number cube letting each of the numbers represent a different shape. (Ignore 6.) 3.Sample: Use a spinner with 8 equal parts. 4. Sample: Roll a 12-sided number cube, letting each of the numbers represent a different symbol. page 103 Reteaching 2: Simulations 1. Sample: Use a spinner divided into four parts, one for each symbol. 2. Sample: Roll two number cubes, letting each of the36 combinations represent a different saying. 3. Sample: Toss a number cube. 4. Sample: Use two sets of number cards 1–10 and draw them from a bag. page 104 Practice 1: Simulations 5 1a. 18 1b. 87 2a. 11 16 b. 16 3. Check students’ work. Answer should be about 12.5%. 4. Check students’ work.Answer should be about 37.5%. 5. Check students’ work. Answer should be about 37.5%. 6. Check students’ work. Answer should be about 12.5%. 7. Possible answer: Let 1 basket made, 2 – 6 basket missed. 8. Check students’ work. page 105 Practice 2: Simulations 1. Sample: Use a number cube withone number for W, I, E, and R, and two numbers for N. page 106 Assessment 7 1. Answers may vary. Sample: Registered drivers selected randomly is best because that represents a fair sample of the Answers 23 Data Analysis and Probability Workbook Answers relevant population, while people at therestaurant, waiting to buy tickets, or visiting the dealership probably do not. 2. Open and fair. The question lets people answer generally rather than having them pick from limited choices, so it is open. It asks their opinion neutrally without suggesting an answer so it is fair. 3. 260 people 4. $20,000 5. Thesimulation shows that even with a 50/50 chance, the outcomes will not be exactly equal. In the simulation, the 75 coin flips come out 41 heads and 34 tails. The number of students is 10 times the number of coin flips, so this is equivalent to 410 students preferring one beverage and 340 wanting the otherdrink (rather than exactly 375 of each). Based on this result, the beverage committee should buy 35 extras of each drink (410 instead of 375) if they want a good chance of being able to satisfy everyone. pages 107–108 Cumulative Assessment 1. Other 7% Transp. 27% Book/CD 33% Food 33% 2. 86.8,88, 88 3. The median, because there is an outlier and there is no mode. 4. Any number greater than or equal to 27. 5. 44 6. One or both of the axes can be stretched or © Pearson Education, Inc. All rights reserved. compressed so that the rate of change looks bigger or smaller than it really is. Or the y-axiscan be broken so that the differences in magnitude are exaggerated. 7. 35, 210 1 1 8. 13,440 9. 9,000 more 10. 256 11. 18 12. 17 4 1 1 1 ;6 13. 221 14. 169 15. 8 16. 160 17. 25 18. Answers will vary. Sample answer: Do you prefer the greasy pizza from Joe’s or the fancier pizza from Pistachios? Re-write: Which pizza do you prefer, Joe’s or Pistachios? 19. 40,000 people 24 Answers Data Analysis and Probability Teacher’s Guide Answers Activity 1 Activity 5 1. Check students’ work. 2. 1 3. 2 4. Category numbers Possible stores for best buy Store with best buy so that each store occurs exactly twiceNumber of twists Number of strips when cut along the center line 1 and 2 A A 1 1 1 and 3 C, D C 2 2 1 and 4 A, C A 3 1 1 and 5 D, E E 4 2 2 and 3 B B 2 and 4 A, B B 2 and 5 D D 3 and 4 C C 3 and 5 D D 4 and 5 E E 5. Sample: A Möbius strip with 5 twists may produce 1 new strip, and one with 6 twistsmay produce 2 new strips. 6. Sample: By examining the information from the experiments, it is likely that Möbius strips with an odd number of twists result in 1 new strip when cut down the middle. However, Möbius strips with an even number of twists result in 2 new strips when cut down the middle.Activity 6 Activity 2 1a. no 1b. There is an infinite number of ways. 2a. no 2b. There is an infinite number of ways. 3. 1 way 4a. Sample: The cardboard is not stable. 4b. There is an infinite number of ways. 5a. Sample: Yes, it is possible. © Pearson Education, Inc. All rights reserved. However, if all fourerasers are not the same length, they cannot all touch the cardboard. 5b. The cardboard is stable. 5c. The cardboard is not stable. 6. Sample: Three distinct points are needed to determine a unique plane. However, if one line and another point are present, a unique plane can be determined. Activity 3 1.Check students’ work. 2. Check students’ work. 3. ABCD, ABFE, EFGH, CDHG, ADHE, BCFG, ABGH, CDEF, BCHE, ADGF, ACGE, BDHF 4. 12 Activity 4 1.–2. Hypothesis (p) T or F Negation (Mp) T or F X Y (6 5) T X not greater than Y (6 5) F X Y (6 5) F X not less than Y (6 5) T X 2 Y (6 2 5) T X = Y (6= 5) F Activity 7 1a. The 50-yd line is a transversal across the players’ paths. The measures of the same-side interior s on the west side of the 50-yd line are 65° for player #21 and 90° for player #13. 1b. Because the sum of the measures of the same-side interior s is less than 180°, the two pathways meetbeyond the 50-yd line. Since player #13 has less distance to travel, he should be able to overtake player #21. 2. Measurements may vary, however, the alternate interior s will not be . 3. No; the alternate * ) interior s formed are not . 4. It was not, because BX is not 6 to the 40-yd line. Activity 8 Triangle 3.IF X > Y THEN PRINT Y, X; IF X < Y THEN PRINT X,Y; IF X = Y THEN PRINT X = Y 4. IF 3 < 10 THEN PRINT 3 = 10 5. False; the hypothesis is true, and the conclusion is false. Geometry Hands-On Activities 1. Sample: HCA and GCF 2. Sample: BEG and IEG, CHG and BHG 3. Sample: DJA and EBA,BAJ and EID 4. Sample: EGH and HGC, DJF and FJA 5. Sample: DJF and HBA, FJA and EBH 6. Sample: DJF and FJA, BAC and CAJ 7. Sample: BEG and BHG 8. Sample: GCH, DIE, and BCA 9. Sample: DGCF, IBAJ, and EGAB 10. EBHG and DJBE Classification IMQ acute, scalene AHD obtuse,scalene BEJ acute, equilateral FLK acute, isosceles GRC right, scalene STP obtuse, scalene Answers 41 Answers (continued) Activity 9 Activity 13 1a. 1b. 1. the width of the straightedge 2. They are equal. 3. They are equal. 4. Point T lies on QS. 5. They are equal. 6. QS is the bisector of PQR in PQR.Activity 14 2. 3. 1. B A 4a. C 2. B 2 D 4b. E J 1 A 3. Activity 10 1.–3. Check students’ work. 4. congruent intersecting circles 5. congruent crescentlike shapes; congruent triangular shapes; congruent oval shapes 6. two pairs of congruent triangles 7. three pairs of congruent triangles 8. Check students’designs for requirements. H 3K F 4 G C 3 1 2 4 4. 1. Because all pins are equidistant, each side length is the sum of two equal distances. So AB JK, BC KL, and AC JL . Therefore, ABC JKL by SSS. 2. TSR VUQ. The pairs of corresponding sides are TS and VU, SR and UQ, and TR and VQ . By the

previous reasoning (each side is the sum of line segments), the pairs of corresponding sides are . So the triangles are by SSS. 3a. Sample: A B C D E F G H I © Pearson Education, Inc. All rights reserved. Activity 11 5. 6. J 3b. Sample: ABE, BFC, CDG, BEF, CFG, EFH, FGI, HIJ, FHI 7. Activity 12 1. Thenumber of smaller triangles in an n-frequency dome is n2. 2. As the frequency increases, the dome becomes more spherical in shape. 42 Answers Geometry Hands-On Activities Answers (continued) Activity 15 Activity 17 Number of Number of differentNumber of isosceles Number of shaped Lengthsequilateral triangles (nontoothpicks triangles of sides triangles equilateral) 3 1 1, 1, 1 1 0 4 0 Not possible 0 0 5 1 2, 2, 1 0 1 6 1 2, 2, 2 1 0 7 2 1, 3, 3 and 2, 2, 3 0 2 8 1 2, 3, 3 0 1 9 3 3, 3, 3; 2, 3, 4; 1, 4, 4 1 1 10 2 2, 4, 4; 3, 3, 4 0 2 11 4 1, 5, 5; 2, 4, 5; 3, 4, 4; 3, 3, 5 0 3 2, 5, 5; 3, 4, 5; 4, 4, 4 1 © PearsonEducation, Inc. All rights reserved. 12 3 1 1. A triangle cannot be formed. 2. 5, 7, 8, 9, 10, 11, 12; there does not seem to be a pattern. 3. 3, 6, 9, 12; yes; the pattern seems to be multiples of 3; prediction: 15. 4. 1, 2, 3, 4; prediction: 5; the prediction is correct: 1, 6, 6; 2, 5, 6; 3, 5, 5; 3, 4, 6; 4, 4, 5. 5.Prediction: 6; the prediction is not correct, 7 triangles can be formed: 1, 7, 7; 2, 6, 7; 3, 5, 7; 3, 6, 6; 4, 4, 7; 4, 5, 6; 5, 5, 5. 6. greater than; cannot Number of toothpicks (y) Sketch of arrangement(s) Number of squares (x) 4 1 7 2 8 1 10 3; 2 11 3 12 1; 5 1. 1 2. 4 3. Sample: One square can be made fromany multiple of 134 toothpicks. 4 1. 1 2. 4 3. Answers may vary. Sample: If the number of toothpicks is divisible by 4, there can be 1 square formed. Activity 16 1. All triangles, when combined with a congruent triangle, form a parallelogram. This is true because any parallelogram split in half on its diagonalproduces two congruent triangles. 2. The two acute isosceles triangles yielded a rhombus. Because of the two equal sides of each triangle, the two triangles form four congruent sides when a pair of sides are placed together. 3. The two right scalene triangles yielded another triangle. This triangle isclassified as isosceles. This is possible because of the 90° angle in each triangle. When joined, these two 90° angles form a straight angle. 4. The two obtuse scalene triangles yielded a concave quadrilateral. When these triangles are joined at their shorter sides, an interior angle greater than 180° isformed. 5. All three pairs of triangles yielded a kite. When the triangles are joined at their longer sides, congruent sides are adjacent to each other, which is the definition of a kite. 6. parallelogram, rhombus 7. parallelogram, rhombus Activity 18 1. square, rectangle, rhombus, or parallelogram 2. square orrectangle 3. square or rhombus 4. quadrilateral European kite European trapezoid European isosceles trapezoid parallelogram rhombus rectangle square Geometry Hands-On Activities Answers 43 Answers (continued) 3b. Sample: 5. Quadrilaterals with area 10 square units general quadrilateralClassification trapezoid MutBis MutBis Dimensions rectangle b = 5, h = 2 parallelogram b = 2, h = 5 isosceles trapezoid b1 = 6, b2 = 4, h = 2 kite d1 = 4, d2 = 5 trapezoid b1 = 3, b2 = 1, h = 5 s = "10 square 4a. Sample: MutBis MutBis Activity 19 1. Check students’ work. 2. Sample: The area of JKLM isone-fifth the area of ABCD. 3.–5. 10 D G 8 2 Classification L 6 4 Quadrilaterals with area 12 square units C M H F K J A E 2 4 6 B 8 10 12 * ) * ) * ) * ) * ) * ) *ED) : y =* -)2x + 10; AF and CH have the same * slopes; ) BG and ED have the same slopes. The slopes of AF and * ) * ) * ) CH are the negativereciprocals of the slopes of BG and ED . * ) * ) 6. AF : y = 12 x, CH : y = 12 x + 5, BG : y = -2x + 20, 7. They are parallel. BG and ED are also parallel. JKLM is a parallelogram. 8. J(4, 2) 9. K(8, 4), L(6, 8), M(2, 6) 10. area of JKLM = 20 square units; area of ABCD = 100 square units; ratio = 15 11. Checkstudents’ work. Dimensions rectangle b = 6, h = 2 rectangle b = 4, h = 3 isosceles trapezoid b1 = 5, b2 = 3, h = 3 kite d1 = 6, d2 = 4 parallelogram b = 3, h = 4 trapezoid b1 = 3, b2 = 1, h = 6 5a. Sample: 5b. Sample: Quadrilaterals with area 15 square units Classification Dimensions Activity 20 rectangle b= 3, h = 5 1a. Check students’ work. 1b. Check students’ work. 2. rectangle: b = "18, h = "2; kite: d1 = 2, d2 = 6; parallelogram b = 5, h = 3 isosceles trapezoid b1 = 4, b2 = 2, h = 5 rectangle b = 15, h = 1 trapezoid b1 = 7, b2 = 3, h = 3 parallelogram: b = 3, h = 2; trapezoid: b1 = 4, b2 = 2, h = 2 3a. Sample:6. Sample: Use factors of the given area. 7. Sample: Work backward from area formulas of parallelograms, trapezoids, and kites. 44 Answers Geometry Hands-On Activities © Pearson Education, Inc. All rights reserved. 12 4b. Sample: Answers (continued) 6. Activity 21 1. 100 square units 2. PS = 6; SG= 8 3. YPS SGA by SAS, so SA YS by CPCTC. 4. 1 4 and 2 3 by CPCTC. 5. They are complementary. 6. 90 7. Check students’ work; area: c2 = 100. 8. original squares: 82 + 62 = 64 + 36 = 100; new squares: 10 ? 10 = 100 9. Check students’ work. 10. X U R a Y c S c V Z 7. and 8. b T Using the figureabove, the area of square XYZU is a2, and the area of square RZST is b2. So the area of figure XYTSRU is a2 + b2. If you cut out XYV, VST, and figure XVSRU and then rearrange them to form a square with sides c, the area of the square formed is c2. Therefore, a2 + b2 = c2. Activity 22 1. 9. 10. 2. ©Pearson Education, Inc. All rights reserved. 3. Activity 23 1. Check students’ work. 2. Check students’ work. 3. Triangle A: hypotenuse = 5 cm; Triangle B: hypotenuse 4. 5. = 13 cm; Triangle C: hypotenuse = 10 cm; Triangle D: hypotenuse = 15 cm; Triangle E: hypotenuse = 17 cm 4. 3, 4, 5; 5, 12, 13; 6, 8,10; 9, 12, 15; 8, 15, 17 5. 3, 4, 5; 6, 8, 10; 9, 12, 15 6. Sample: Each side of a bigger triangle is a multiple of its corresponding side of a smaller triangle. For triangles to be similar, they must have a common ratio among their sides. 7. Sample: 10, 24, 26; 16, 30, 34 8. Sample: similar to triangle B: 10, 24, 26;15, 36, 39; similar to triangle E: 16, 30, 34; 24, 45, 51 9a. 72 + 242 = 252; 49 + 576 = 625; 625 = 625 9b. Sample: 14, 48, 50; 21, 72, 75 Activity 24 1. Check students’ work. 2. ABC Geometry Hands-On Activities mA = 90 mB = 27 mC = 63 MNP mM = 90 mN = 27 mP = 63 RST mR = 90 mS = 27 mT = 63ABC AB = 3 cm BC = 3.4 cm CA = 1.5 cm MNP MN = 4 cm NP = 4.5 cm PM = 2 cm RST TR = 2.5 cm RS = 5 cm ST = 5.6 cm Answers 45 Answers (continued) RST, and the sides of MNP are proportional to the sides of RST. 4. Yes; by Theorem 8-2, if the ratios are equal, then the triangles are similar. 5a.They are parallel. 5b. MW is half the length of MP. MX is half the length of MN. 6. The length of the part of the shorter leg is half the length of the part of the longer leg. Activity 25 1. Check students’ work. 2. Triangle A: hypotenuse 3.6 cm; Triangle B: hypotenuse ≈ 7.2 cm; Triangle C: hypotenuse 14.4 cm3. The corresponding sides are proportional. 4. The three triangles are similar. 5. The corresponding angles are congruent. 6. Smaller angle side opposite side adjacent Triangle A 2 3 1 1.8 1 1.2 Triangle B 2 3 1 1.8 1 1.2 Triangle C 2 3 1 1.8 1 1.2 side opposite side adjacent hypotenuse hypotenuse 7. Theratios in each column are equal. 8. Larger angle side opposite side adjacent Triangle A 3 2 1 1.2 1 1.8 Triangle B 3 2 1 1.2 1 1.8 Triangle C 3 2 1 1.2 1 1.8 side opposite side adjacent hypotenuse hypotenuse The values in the second column are the reciprocals of the values in the second column of thetable in Exercise 6. The values in the third and fourth columns are switched with the values in the third and fourth columns of the table in Exercise 6. Activity 27 1. 180 2. 270 3. 315 4. 112.5 5. The paper moves faster than the speed of the force with which you rolled it. Its actual speed is the sum of theforce with which you rolled the paper and the force with which you blew on it. Blow Push 6. The paper moves slower than the speed of the force with which you rolled it. Its actual speed is the difference between the force with which you rolled it and the force with which you blew on it. Blow Push 7. BlowActual path Push 8. The faster you roll the paper, the more closely it follows its original direction. The harder you blow, the more the paper deviates from its original direction. 9. 580 mi/h 10. 420 mi/h 11. Actual velocity Wind 80 mi/h Velocity with no wind 500 mi/h 12a. 506 mi/h 12b. This result is betweenthe speed in a headwind and the speed in a tailwind. This makes sense because a headwind gives the greatest resistance, and a tailwind gives the greatest assistance. 13a. 9° 13b. This makes sense when compared with the vector diagram that illustrates the situation (see answer to Exercise 11). Thecourse of the plane is slightly north of east. Activity 28* ) * ) * ) 1. Sample: G–AB –D, A– CD * –E, ) C– EG –A * ) GD –E 4. A– EC –G 2. triangular prism 3. B– * ) 5. E– CD –A 6. AEC and BGD Activity 26 1.–4. Check students’ work. 46 Answers Geometry Hands-On Activities © Pearson Education, Inc. Allrights reserved. 3a. These sides are approximately proportional. 3b. yes 3c. Yes; the sides of ABC are proportional to the sides of Answers (continued) 1.–2. Check students’ work. 3. The angles appear to be supplementary. 4. Yes; the angle formed by the Activity 29 1. Check students’ work. 2. Circle 1 23 Radius of circle (R) Take-out angle (x) Cone angle (y) A 10 cm 72 288 B 10 cm 90 270 C 10 cm 180 180 Activity 32 D 10 cm 216 144 1., 3.–6., and 8. 4 5 6 cone angle 360 Radius of base of cone (r) Circle Key x cone radius circle radius 4 5 8 cm B 3 4 15 cm 2 3 4 C 1 2 5 cm 1 2 D 2 5 4 cm 2 5 A ©Pearson Education, Inc. All rights reserved. two tangents and the angle formed by the two radii are supplementary. 5. APO BPO. Justifications will vary, but the triangles must be congruent by the HL Theorem (right angles by construction, PO is the hypotenuse of both triangles, and the congruent radii OAand OB are the corresponding legs). 6a. PO is the bisector of APB by CPCTC. 6b. PO is the bisector of AOB by CPCTC. 4 5 3. The radius of the circle is the slant height of the cone. 4. Slant height of cone 5. The values are equal. 2 x 2 x 6. 2pr = 360360 (2pR) 7. r = (360360 )R 360 2 x r 8. R = 360 9.circle A: h = 6 cm; circle B: h 6.6 cm; circle C: h 8.7 cm; circle D: h 9.2 cm 10. Check students’ work. F M O Altitudes Medians Perpendicular bisectors F M C2 x x C1 F M Euler line C1 – Circumcenter O – Orthocenter C2 – Centroid x – Euler Points F – Feet of Altitudes 2. The distance from the centroid tothe orthocenter is twice the distance from the centroid to the circumcenter. 7. The center of the nine-point circle lies on the Euler line and is midway between the circumcenter and the orthocenter. 9. The radius of the nine-point circle is equal to half the radius of the circumscribed circle of the triangle.Activity 33 Activity 30 1. Check students’ work. 2. 4 3. They are congruent. 4. 6 5. no 6. 2 7. The sum will always be greater than 180º. 8. Sample: Any two points on a sphere lie on the same great circle. Therefore, the shortest distance between any two locations on Earth will be along a great-circle route.1. 90 2. 180 3. semicircle 4. Sample: AG = 42 mm; BG = 42 mm; CG = 42 mm; DG = 42 mm 5. G is the center of the circle. 6. Check students’ work. Activity 34 1. Sample: Activity 31 Steps 1.–5. 1 of base 2 P C 2. Sample: D A E B base O T Geometry Hands-On Activities Answers 47 Answers (continued)3. Sample: 8. altitude of k endpoint of base endpoint of base 4. Sample: 9. perpendicular bisector 10. Locate the perpendicular 2 of base 3 bisector of the base of each triangle, and then reflect, or draw a mirror image of, the triangle on each side of this line. 11. Check students’ work. Activity 35 1. Figures4 and 6 2. Figures 3 and 1 3. Figure 5 4. Figure 2 5.–8. Check students’ work. 5. side Activity 36 1. Sample: part of base 6. base part of side 2. 60 3. 40 4. 5 5. 2 6. 90; 65; 5; 2 7. Check students’ work. 7. vertex 48 Answers © Pearson Education, Inc. All rights reserved. endpoint of base Geometry Hands-On Activities Answers Screening Test Quarter 1 Test, Form A 1. C 2. H 3. B 4. H 5. C 6. J 11. C 12. F 13. B 14. H 15. C 19. B 20. J 21. C 22. H 23. C 27. D 28. G 29. D 30. F 31. C 35. C 36. G 37. D 38. F 39. B 7. A 8. F 9. C 10. G 16. J 17. D 18. H 24. J 25. B 26. H 32. F 33. C 34. J 40. J 41. C 42. J 1. 13,21 2. Sample: a tiger 3. right triangle 10 in. 4. 6 in. Benchmark Test 1 All rights reserved. 1. D 2. H 3. B 4. F 5. C 10. F 11. B 12. H 13. D 17. B 18. G 19. C 20. F 24. G 25. A 26. H 27. D 31. B 32. F 33. C 6. H 7. D 8. J 9. A 14. G 15. D 16. F 21. C 22. G 23. C 28. G 29. C 30. J 2 in. 10 in. 2 in. 5. -27 6. x = 37. 25 8. L M Benchmark Test 2 1. A 2. H 3. A 4. J 5. A 6. H 7. B 8. F 9. C 10. H 11. A 12. G 13. C 14. G 15. B 16. H 17. B 18. H 19. B 20. F 21. C 22. G 23. D 24. H 25. B 26. J 27. C 28. G 29. A 30. H 9. Subtraction Property 10. Benchmark Test 3 © Pearson Education, Inc., publishing as Pearson PrenticeHall. 1. A 2. F 3. 10. G 11. A 17. B 18. H 24. H 25. B 31. B 32. H D 4. J 5. B 6. F 7. B 8. F 9. D 12. G 13. C 14. F 15. B 16. G 19. D 20. H 21. C 22. H 23. D 26. J 27. C 28. H 29. C 30. G 33. B Benchmark Test 4 1. B 2. F 3. C 4. J 5. A 10. G 11. B 12. H 13. C 17. B 18. J 19. D 20. G 24. F 25. D 26. F 27. A 6.F 7. C 8. 14. J 15. C 21. B 22. H 28. F 29. B J 9. D 16. J 23. B 30. H Benchmark Test 5 1. B 2. H 3. B 4. J 5. C 6. F 7. B 8. J 9. D 10. F 11. C 12. H 13. B 14. H 15. A 16. G 17. D 18. F 19. C 20. H 21. B 22. H 23. A 24. H 25. B 26. J 27. B 28. F 29. D 30. G 31. D 32. J 33. B 34. G 35. A 36. J 37. D 38. J 39. CAnswers Fro nt ht Rig 11. 70 12. 62 13. 29 14. Suppose that 2x + 4 = 6. Subtract 4 from both sides of the equations by the Subtraction Property. This gives an equation of 2x + 4 - 4 = 6 - 4, which reduces to 2x = 2. Now divide both sides of the equation by 2 by using the Division Property. This gives an * )2 equation of 2x 2 = 2 , which simplifies to x = 1. 15. AC 16. 12p m 17. Sample: A student can drive if and only if he or she is over the age of sixteen. If a student can drive, then he or she is over the age of sixteen. If a student is over the age of sixteen, then he or she can drive. 18. 144 19. (1, 2.5) 20. -2321. 2 4 by the Converse of the Corresponding Angles Theorem or 2 + 3 = 180 by the Converse of the Same-Side Interior Angles Theorem 22. 2 "17 23. no intersection or a point 24. Two lines must not intersect and be in the same plane to be parallel. 25. 40 26. 50 27. 90 28. 75 29. hypothesis: if you arenot at school; conclusion: you are at home 30. If you are at home, then you are not at school. 31. 20 in. 32. Transitive Property 33. center (5, -5); radius = "5 Geometry 155 Answers (continued) Quarter 1 Test, Form B Quarter 1 Test, Form D 1. 8, 5 2. Sample: obtuse triangle 3. isosceles triangle 3 cm 4. 14cm 1. 21, 25 2. scalene triangle 3. 36p m2 4. A 5. 5 6. 9 7. 14 cm 6 cm 1 Right 2 3 cm Front 8. 138 9. 46 10. C 11. 40 cm 12. BF 13. 10p m 14. 120 15. (-2, 4) 16. 12 17. B 18. 5 19. 45 20. 90 21. 60 22. 85 23. hypothesis: two angles have a sum of 5. -18 6. x = 6 7. 7 8. 3 A B 180; conclusion: the angles aresupplementary angles 9. Addition Property Quarter 1 Test, Form E 10. 1. 26, 31 2. equilateral triangle 3. 100p m2 4. 5 5. 6 6. 2 Fro nt h Rig t 11. 112 12. 60 13. 44 14. Suppose that 3x + 5 = 11. Subtract 5 from both sides of the equation by the Subtraction Property. This gives an equation of 3x + 5 - 5 = 11- 5, which reduces to 3x = 6. Now divide both sides of the equation by 3 by using the Division Property. This gives an 6 equation of 3x 3 = 3 , which simplifies to x = 2. 15. point R 16. 26p m 17. Sample: A student can drive if and only if he or she is over the age of sixteen. If a student can drive, then he orshe is over the age of sixteen. If a student is over the age of sixteen, then he or she can drive. 18. 135 19. (-6, 11) 20. 32 21. Sample: 1 3 or 2 4 by the Converse of the Corresponding Angles Postulate; 2 + 3 = 180 by the Converse of the Same-Side Interior Angles Theorem 22. 5 23. no intersection or aline 24. A triangle cannot have two sides and no sides equal at the same time. 25. 60 26. 90 27. 70 28. 45 29. hypothesis: if you are not at work; conclusion: you are at school 30. If you are at school, then you are not at work. 31. 96 in. 32. Substitution Property 33. center (2, 1); radius = "117 156 Geometry2 2 Right 1 Front * ) 7. 110 8. 70 9. B 10. G 11. 60 in. 12. AB 13. 14p m 14. 135 15. (1, 5) 16. -13 17. D 18. 10 19. 72 20. 55 21. 50 22. 75 23. G 24. Hypothesis: two lines form right angles; conclusion: the lines are perpendicular. 25. C Quarter 2 Test, Form A 1. RS 2. 20 3. trapezoid 4. 50 5. 40; 82 6. SSSPostulate 7. SAS Postulate 8. (c - a, b) 9. Sample: (0, 4), (4, 0), (-4, 0), (0, -4) 10. R, P, Q 11. AAS, ABC ABD 12. 133; 47 13. Sample: 4 ft, 3 ft, and 8 ft 14. 4 15. Sample: The diagonals must be perpendicular bisectors of each other, or all sides must be congruent. 16. The orthocenter is the intersection ofthe three altitudes of a triangle. 17. an altitude 18. 45 19. Hypotenuse-Leg Theorem 20. centroid 21. It is the median and altitude of the triangle. 22. Suppose that a triangle had more than one right angle. If this were true, then two of the angle measures alone would add up to 180, and the third angle wouldhave a measure that would contradict the Triangle Angle-Sum Theorem. 23. AB , BC , AC 24. 90; 30 25. rhombus 26. 65 27. c, e, b, a, d or e, c, b, a, d 28. No; the midsegment is not half the length of the third side. 29. a kite 30. 31 31. They must be perpendicular bisectors of each other. 32. AC Answers ©Pearson Education, Inc., publishing as Pearson Prentice Hall. C All rights reserved. 24. G 25. D Answers (continued) Quarter 2 Test, Form B All rights reserved. 1. B 2. 10 3. parallelogram 4. 58 5. 41; 75 6. ASA Postulate 7. AAS Theorem 8. (c + a, b) 9. Sample: (0, 4), (8, 0), (-4, 0), (0, -4) 10. L, N, M 11.SAS, ACD CAB 12. 142; 38 13. Sample: 4 ft, 3 ft, and 8 ft 14. 4 15. Sample: The diagonals must be perpendicular bisectors of each other and all sides must be congruent, or all sides and angles must be congruent. 16. The incenter is the intersection of the three angle bisectors of a triangle. 17. a median18. 41 19. Hypotenuse-Leg Theorem 20. centroid 21. It is the median and altitude of the triangle. 22. Suppose that a triangle had more than one obtuse angle. If this were true, then two of the angle measures alone would add up to more than 180, and the third angle would have a measure that wouldcontradict the Triangle Angle-Sum Theorem. 23. AB , BC , AC 24. 90; 50 25. rectangle 26. 70 27. a, c, e, b, d or c, a, e, b, d 28. Yes; the midsegment is half the length of the third side. 29. a parallelogram 30. 31 31. They must be equal in measure. 32. AC + BC © Pearson Education, Inc., publishing asPearson Prentice Hall. Quarter 2 Test, Form D 1. XZ 2. 4 3. D 4. 55 5. x = 85, y = 60 6. J 7. C, A, B 8. m1 = 54.5, m2 = 54.5, m3 = 71 9. D 10. C 11. B 12. E 13. 55 14. 56 15. H 16. m1 = 55, m2 = 45, m3 = 45, 17. m1 = 25, m2 = 25, m3 = 130 18. A 19. parallelogram, rectangle, square, rhombus 20. 164 21.B D 22. altitude 23. 33 25. Sample: 6 4 6 Quarter 3 Test, Form A 1. 14 2. ABC JKL by SSS ~ Theorem 3. TUV WXY by AA ~ Postulate 4. obtuse 5. 56.3 6. 10.7 7. reflectional symmetry and 180º rotational 8 symmetry (point symmetry) 8. 2 "26 9. sin A = 17 ; 8 15 cos A = 17 ; tan A = 15 10. A(7, 2), B(11,2), C(10, -1), D(6, -1) 11. A(-5, 4), B(-9, 4), C(-8, 1), D(-4, 1) 12. A(-3, -4), B(-7, -4), C(-6, -1), D(-2, -1) 13. 150 in.2 14. 53.9 mi/h; 21.8° south of east 15. A 16. (3, 3) 17. 4 "3 18. 38 19. 6 20. 18.4 21. Check students’ work. 22. No; there will be gaps when the pattern is repeated. 23. A(-2, 6), B(0, 14), C(8, 4)24. 48 ft Answers 9 6 6 9 26. x 5 3, y 5 3"2 27. x = 2, y = 4 28. 3, -9 29. 12 in. Quarter 3 Test, Form B 1. 14 2. not similar 3. ABC LPR by SAS ~ Theorem 4. obtuse 5. 14.9 6. 55.0 7. reflectional symmetry 8. "66 9. sin A = 53; cos A = 54; tan A = 34 10. A(-6, -3), B(-2, -3), C(-2, 0), D(-5, 0) 11. A(-18, 15), B(-6, 15), C(-6, 6), D(-15, 6) 12. A(-9, 6), B(-5, 6), C(-5, 3), D(-8, 3) 13. 18 cm2 14. 40.3 mi/h; 29.7º north of west 15. D 16. (-4, 1) 17. 2 "34 18. 4 19. 63 20. 33.7 21. Check students’ work. 22. yes 23. X(0, -3), Y(3.5, -2), Z(3, 1) 24. 265 ft 25. Sample: 3 Quarter 2 Test, Form E 1. R 2. 5 3. C 4. 50, 65 5. x = 100,y = 80 6. F 7. C, B, A 8. m1 = 122, m2 = 29, m3 = 29 9. E 10. A 11. C 12. B 13. 78 14. 58 15. G 16. m1 = 44, m2 = 84, m3 = 40 17. m1 = 65, m2 = 65, m3 = 50 18. D 19. parallelogram, rectangle, square, rhombus 20. 187 21. BX DX 22. perpendicular bisector 23. 40 4 3 6 6 26. x 5 5, y 5 5"2 27. x = 3, y = 628. 2, 9 29. 35 in. Quarter 3 Test, Form D 1. BRG NDK by SAS Theorem 2. AYM XQH by AA Postulate 3. 12.8 5 4. 32.2 5. 13.5 6. D 7. 4 "5 8. sin A = 13 ; cos A = 5 2 tan A = 12 ; 9. (10, 0) 10. G 11. 112 in. 12. A 12 13 ; 13. 5"7 14. x = 9, y = 15 15. y = 14 16. X(4, 6), Y(5, 10), Z(8, 7) 17. X(2, -1), Y(3, -5),Z(6, -2) 18. X(-2, -1), Y(-3, -5), Z(-6, -2) 19. 6"3 20. 88 ft 21. 12 in. 22. J Quarter 3 Test, Form E 1. WLB HER by AA Postulate 2. CMJ FKP by SAS Theorem 3. 24 4. 12.2 15 8 5. 63.0 6. C 7. 6"3 8. sin A = 15 17; cos A = 17; tan A = 8 ; Geometry 157 (continued) 9. (-6, 9) 10. J 11. 588 in.2 12. A 13. 8"1014. x = 3, y = 25 15. x = 10 16. A(0, 1), B(-3, 7), C(-4, 2) 17. A(4, -2), B(7, 4), C(8, -1) 18. A(-1, 0), B(-7, 3), C(-2, 4) 19. 9"3 20. 124 ft 21. 3 in. 22. F Quarter 4 Test, Form A 1. 268.1 cm3 2. 66.3 cm2 3. 25 4. 32 5. 144 6. Sample: Draw d1, a diagonal of the kite that divides it into two congruent triangles. Letd1 represent the base of each triangle. The area of one triangle is 21 d1h1. The area of the 14. 122 15. 91.1 16. 75.9 cm3 17. 2 cans of paint 18. 55 19. 35 20. 270 21. 11 22. 93.6 in.2 23. b, c, a, d 24. (x + 3)2 + (y + 2)2 = 9 25. 423.2 cm 2 26. 73 p ft 0 0 27. Sample: RS is an arc of Q. The measure of RSis equal to the measure of RQS. 28. 7162.8 mm2 Quarter 4 Test, Form D 1. 904.8 in.2 2. 4 : 7, 16 : 49 3. 90 4. 75 5. B 6. 96 cm2 7. 8 8. 245.0 ft2 9. 40 cm2 10. 5 11. 9 12. center: (0, 0); radius: 5 y other triangle is 12 d1h2. Because the triangles are congruent, 4 h1 = h2. The other diagonal, d2, is the sumof h1 and h2. Therefore, the area of a kite is 12 d1d2. 7. 14.14 cm2 8. 12.5 9. a circle with center (-2, -3) and radius 5 units 10. 113.1 m2 11. 81 m2 12. 6 13. center: (5, 6); radius: 4 2 2 4 x 4 2 2 4 2 4 y 13. 104 14. 90 15. 128 cm2 16. 184 in.2 17. 180 cm2 18. 201.1 ft2 19. 133 20. 180 21. 192.5 in.2 22.150.80 m 2 23. G 24. 188.5 cm2 x 2 4 6 8 10 Quarter 4 Test, Form E ft3 14. 73.7 15. 92 16. 400 17. 15 rolls of wallpaper 18. 30 19. 60 20. 270 21. 8 22. 310.4 in.2 23. b, c, a, d 24. (x - 2)2 + (y - 2)2 = 4 25. 188.1 in.2 26. 2p in. * ) * ) 27. Sample: QR is tangent to P. QR is perpendicular to a radius of P. 28.284.7 cm2 1. 1436.8 cm3 2. 3 : 8, 9 : 64 3. 65 4. 140 5. B 6. 255 in.2 7. 16 8. 326.7 in.2 9. 95 m2 10. 10 11. 15 12. center: (0, 0); radius: 4 6 y 2 Quarter 4 Test, Form B 1. 1436.8 in.3 2. 63.4 in.2 3. 39.5 4. 25.5 5. 108 6. Sample: Draw d1, a diagonal of the rhombus that divides it into two congruenttriangles. Let d1 represent the base of each triangle. The area of one triangle is 21 d1h1. The area of the other triangle is 12 d1h2. Because the triangles are congruent, h1 = h2. The other diagonal, d2, is the sum of h1 and h1. Therefore, the area of a rhombus is 12 d1d2. 7. 19.63 in.2 8. 6 9. a circle withcenter (-4, 6) and radius 8 units 10. 466.5 ft2 11. 42 in.2 12. 12 13. center: (-3, 5); radius: 3 y 8 6 4 2 64 2 2 4 6 158 x 2 4 6 Geometry x 2 2 2 13. 90 14. 87 15. 125 cm3 16. 804.2 m2 17. 483.8 in.2 18. 262 ft2 19. 59 20. 239 21. 173.8 in.2 22. 94.25 cm2 23. H 24. 55.0 in.2 Mid-Course Test, Form A 1. 63;127 2. 86 3. 12 4. 7 5. rectangle 6. 12 7. Sample: (2, 0), (8, 0), (10, 5), (4, 5) 8. B, A, C 9. 36, 36 10a. If it is summer, then it is sunny. 10b. If it is not sunny, then it is not summer. 11. C 12. 26 13. 68 cm 14. 60 15. CAB BDC by SAS. 16. (0, b) 17a. 3 and 5 or 4 and 6 17b. 1 and 5, 2 and 6, 3 and 7, or 4 and8 18. 90; 20 Answers © Pearson Education, Inc., publishing as Pearson Prentice Hall. 10 8 6 4 2 2 All rights reserved. Answers Answers (continued) 20. G 21. m1 = 105, m2 = 75, m3 = 105, m4 = 75 22. parallelogram, rectangle, rhombus, square 23. B 24. Hypothesis: a transversal intersects two parallellines; conclusion: alternate interior angles are congruent. 25. J 26. A 27. 166 28. y 19. 3 1 Right 2 Front 4 All rights reserved. 20. 115; 65 21. always 22. never 23. never 24. sometimes 25. x = 34; y = 24; z = 88 26. Sample: A rectangle always has opposite sides parallel, making it a parallelogram. Aparallelogram doesn’t always have four right angles, so it is not always a rectangle. 27. C, A, B 28. rectangle 29. rhombus 30. square 31. trapezoid 32. rectangle 33. rhombus 34. 17 35. 48; 90; 42; 57; 33 36. x = 25; y = 31 37. (2.5, 1) 38. J 39. 118; 34 40. 7.2 41. area = 452.4 in.2; circumference = 75.4 in.42. 74 43. HL 44. not possible 45. SSS 46. AAS 47. not possible 48. SAS 49. ASA or AAS 50. SSS or SAS x 4 2 2 4 2 4 2 4 29. 4 y 2 Mid-Course Test, Form B 1. 22, 29 2. 120 3. 17 4. 12 5. parallelogram 6. 7 7. Sample: (0, 4), (-6, 0), (0, -4) and (6, 0) 8. B, C, A 9. x = 120; y = 30 10a. If the ground is wet,then it is raining. 10b. If it is not raining, then the ground is not wet. 11. D 12. 12 13. 120 m 14. 36 15. WXY XWZ by AAS. 16. (a - b, c) 17a. 2 and 3 or 6 and 7 17b. 1 and 3, 2 and 4, 5 and 7, or 6 and 8 17c. 2 and 6 or 3 and 7 18. 34; 68 19. © Pearson Education, Inc., publishing as Pearson Prentice Hall. 22 2 2 x 4 2 2 4 30. 42 31. (1, 2) 32. 120 cm 33. 13 34. G 35. 24 36. 95 37. 157 38. DX JX 39. m1 = 40, m2 = 95, m3 = 45 40. m1 = 54, m2 = 63 Right 1 Front 20. 63; 117 21. always 22. never 23. sometimes 24. sometimes 25. x = 30; y = 36; z = 84 26. Sample: Because m1 + m2 = 135 and m2 + m3 = 135,m1 + m2 = m2 + m3 by substitution. By the Subtraction Prop. of Equality, m1 = m3, and therefore 1 3. 27. YZ, XZ, XY 28. quadrilateral 29. kite 30. trapezoid 31. parallelogram 32. rhombus 33. square 34. 5 35. x = 49; y = 90; z = 18 36. w = 40; x = 30; y = 30; z = 90 37. (1, -6) 38. J 39. 71; 34; 75 40. 14.441. area = 254.5 cm2; circumference = 56.5 cm 42. 54 43. ASA 44. SSS 45. SAS and ASA 46. not possible 47. HL 48. not possible 49. AAS 50. SAS Mid-Course Test, Form E 1. -19, -23 2. 13 3. 12 4. RT , RS, TS 5. B 6. 24 7. F 8. 60 9. C 94.2 in., A 706.9 in.2 10. 13 11. m1 = 47, m2 = 43 12. AAS 13.ASA 14. SAS 15. not possible 16. HL 17. SSS 18. 50 19. D 20. J 21. m1 = 80, m2 = 80, m3 = 100, m4 = 80 22. square, rhombus 23. A 24. Hypothesis: a transversal intersects two parallel lines; conclusion: corresponding angles are congruent. 25. H 26. B 27. 216 28. 4 y 2 x 4 2 2 4 2 Mid-Course Test,Form D 1. -9, -11 2. 11 3. 16 4. T, R, S 5. D 6. 14 7. G 8. 45 9. C 100.5 in., A 804.2 in.2 10. 10 11. m1 = 38, m2 = 52 12. SAS 13. HL 14. not possible 15. SSS 16. AAS 17. ASA 18. 19 19. A Answers Geometry 159 Answers 29. 4 (continued) 27. A, C, B 28. 17.9 29. never 30. sometimes 31. always 32.sometimes 33. never 34. 18 35. 34 36. 92.3 in.2 37. 100 38. 25 : 4 39. 199.0 cm2 40. 9.7 41. y 2 x 2 2 4 2 42. 59 55.6% 43a. 1809.6 cm2 43b. 7238.2 cm3 44a. If 4 30. 52 31. (2, 1) 32. 160 cm 33. 23 34. J 35. 47 36. 80 37. 67 38. MZ TR or WZ WR 39. m1 = 52, m2 = 30, m3 = 98 40. m1 = 59, m2 = 62Final Test, Form A 1. 36 2. (2.5, 1) 3. B 4. G 5. (6, -5) 6. 128 cm2 7. 48 cm2 8. reflectional and rotational symmetry 9. D 12 cm 10. two angles have the same measure, then they are congruent. 44b. If two angles are not congruent, then they do not have the same measure. 45a. PQ 45b. F 46. 13.0 47. AAS48. rhombus 49. 113 50. y = -13 x + 10 Final Test, Form B 1. 45 2. (9, 2) 3. C 4. J 5. (-1, -1) 6. 216 "3 cm2 7. 96 in.2 8. rotational and reflectional symmetry 9. B 10. surface area: 84 cm2 5 cm 6 cm 5 cm 5 cm All rights reserved. 4 10p p cm 11. 13.6 ft 12. Fro nt Rig ht 13. 47.0 cm2 14. 1280 m3 15. Thelocus would consist of Front Right 13. 120 cm2 14. 1296 cm2 15. two parallel, horizontal lines, one 3 units above and one 3 units below y = -2 y 5 4 2 2 4 x a congruent segment 5 cm above and another 5 cm below the given segment. The endpoints of these two segments would be connected withsemicircles of radius 5 cm. 16. 144 "3 in.2 17. 8.9 18. H 19. 6.4 20. 10.5 21. 175 cm2 22. 63 m 23. 44 m 24. 53.2 mi/h; 41.2° west of north 25. 11 26. (-1, -6) 27. C, B, A 28. 10.8 29. always 30. sometimes 31. never 32. sometimes 33. sometimes 34. 5.5 35. 32.5 36. 24.6 cm2 37. 80 38. 81 : 25 39. 182.8cm2 40. 6 41. y 2 2 4 16. 40 m2 17. 15.0 18. F 19. 6.5 20. 7.5 21. 54 cm2 22. 110 ft 23. 64 ft 24. 9, 1 25. 7 26. (-2, -2) 160 Geometry Answers © Pearson Education, Inc., publishing as Pearson Prentice Hall. 11. 14.0 in. 12. Answers (continued) 42. 41 = 25% 43a. 282.7 m2 43b. 314.2 m3 44a. If two anglesare congruent, then they are vertical. 44b. If two angles are not vertical, then they are not congruent. 45a. AE 45b. B 46. 17.3 47. HL 48. square 49. 125 50. y = -12 x - 2 Final Test, Form D triangle are congruent; conclusion: the angles opposite those sides are also congruent 4. H 5. A 6. 62 7. 74 8. 91 9. G10. B 11. C 12. A 13. D 14. A 380.1 in.2, C 69.1 in. 15. (-3, -4) 16. y = -x + 1 17. AAS Theorem 18. ASA Postulate 19. not possible 20. SAS Postulate 21. HL Theorem 22. x = 8, y = 4"3 23. F 24. 200p cm3 25. 40 26. B 27. 90 28. 100 m2 29. 11 ft 30. 17 31. H 32. 28 33. 71 34. 180 35. acute 36. center: (-4,3); radius: 7 37. AC 41.3, CD 33.4 38. 288 square units 39. 8 40. 43.30 cm2 1. x = 18, y = 147 2. D 3. hypothesis: if a point is on the perpendicular bisector of a segment; conclusion: it is equidistant from the endpoints of the segment 4. J 5. A 6. 150 7. 78 8. 76 9. G 10. A 11. D 12. C 13. B 14. A 907.9 m2,C 106.8 m 15. (6, -2) 16. y = -x - 8 17. SAS Postulate 18. not possible 19. SSS Postulate 20. HL Theorem 21. AAS Theorem 22. x = 12, y = 6"3 23. F 24. 540p ft3 25. 8 26. D 27. 180 28. 128 cm2 29. 14 ft 30. 25 31. G 32. 66 33. 68 34. 174 35. obtuse 36. center: (-6, 5); radius: 8 37. MW 43.5, MX 34.3 38.80 39. 15 40. 173.21 in.2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. 1. x = 19, y = 135 2. D 3. hypothesis: if two sides of a Final Test, Form E Answers Geometry 161 (continued) Test-Taking Strategies Interpreting Data 1. C 2. G 3. C 4. J Writing GriddedResponses 1. 6 2. 5 3. 11.4 4. 326.7 5. 615.75 6. 2.5 7. 12 8. 51 9. 87 10. 690 11. 552.9 12. 13.30 Using a Variable is found,all work is shown, and a diagram is complete and correct.2c. 1 point; correct answer, but did not complete a drawing.3a. 0 points; incomplete and incorrect response 3b. 1 point;work is shown, however it contains a math error Writing Extended Responses 1. No steps are shown and no justification is made about why m/1 = m/2. 2. B 4x 30 A C F Answers may vary: Sample: BD bisects /ABC, therefore m/CBD = m/ABD. 2x = 4x - 30 x = 15 m/CBD = 30 m/BDC = 90 The measure ofan exterior angle is the sum of the two remote interior angles therefore the m/BCF = m/CBD + m/BDC = 30 + 90 = 120. 3. Answers may vary. Sample: Find the slope of the line. m 5 1 2 3 5 2 2 5 21 . 6 3 3 2 (23) Parallel lines have the same slope. y 1 1 5 21 (x 1 1) Substitute m = -13 and (-1, -1) 3 in thepoint-slope formula. y 1 1 5 21 x 2 1 3 3 4 1 Solve for y to write the equation in y52 x2 3 3 slope-intercept form. (3, 3) y 2 (3, 1) x 2 O (1, 1) 2 162 Finding Multiple Correct Answers 1. C 2. J 3. C 4. F 5. B 6. G 7. C Testing Multiple Choices 1. B 2. G 3. C 4. F 5. B 6. J 7. C 8. G 9. B 10. F EliminatingAnswers 1a. Answers may vary. Sample: D can be eliminated because 2x D Drawing a Diagram 1. rhombus 2. BE = ED = 8 and AE = EC = 24 3. Two side are 22 and two sides are 14 4. Y(-8, 2); Z(2, 6) 5. 35, 35 6. 79, 79 7. (0, 1), (3, 0) 8. (5, 5) 9. (-6, -6), m = 1 10. 12 1 6"2 or 20.49 11. x = 1; eachdiagonal is 8 12. 120° 13. Two sides are 12 and two sides are 32 All rights reserved. Writing Short Responses 1a. 2 points; all parts are solved, work is shown, and answers are correct. 1b. 0 points; no work shown and answer is incorrect. 1c. 1 point; error in procedure, but problem is set up correctly.incomplete answer 2a. 0 points; incorrect and incomplete response 2b. 2 points; the measure of each angle 7 , 0b 1. 108, 78, and 94 2. 40, 60, and 80 3. (6.5, 0) 4. a2 19 5. 8 or 12 6. 40 7. 60 8. 2212 9. 4 in. 10. 12 11. 3 12. 7 13. 4 14. 108° 2 Geometry 142 = 196 and the length of the side will be less than14. A can be eliminated because the length of the side will be longer than 7. 1b. C 2a. The angle is just a bit larger than a 30° angle, so using the special right triangle relationships you can eliminate F. The hypotenuse is the longest side so you can eliminate J. 2b. G 3a. The angle is just a little bit smallerthan a 30° angle, so using the special right triangle relationships you can eliminate A and D. 3b. B 4a. H and J are obtuse angles. 4b. G 5a. A and B are too small since the area will be greater than 12 ? 7 ? 48. 5b. C 6a. The first coordinate has to be -9 so choices G and J can be eliminated. 6b. FChoosing “Cannot Be Determined” 1. C 2. G 3. D; need lateral surface area or total surface area. 4. G 5. B 6. J; need the dimensions of the garden. 7. A 8. J; need the radius. 9. C 10. G 11. D; need the length of the edge of the base. 12. G Using Estimation 1. A 2. F 3. C 4. G 5. A 6. H Answering theQuestion Asked 1. the sum of the x- and the y-coordinates of the reflected point; D 2. the y-coordinate of the reflected point; J 3. the distance of the translated point to the x-axis; B 4. the vector of the translation; G 5. the sum of the x- and y-coordinates of the rotated point; D 6. the difference between the y-coordinate and the x-coordinate of the image after translation; H 7. The length A'B'; B 8. lines of symmetry in the figure; H 9. a false property of a reg. hexagon; C 10. image of (8, -5); G 11. the equation of the line after reflection; B 12. the image after the dilation; H Answers © Pearson Education, Inc.,publishing as Pearson Prentice Hall. Answers Answers: SAT/ACT Practice Test All rights reserved. Multiple Choice 1. A B C D E 12. A B C D E 23. A B C D E 2. A B C D E 13. A B C D E 24. A B C D E 3. A B C D E 14. A B C D E 25. A B C D E 4. A B C D E 15. A B C D E 26. A B C D E 5. A B C D E 16.A B C D E 27. A B C D E 6. A B C D E 17. A B C D E 28. A B C D E 7. A B C D E 18. A B C D E 29. A B C D E 8. A B C D E 19. A B C D E 30. A B C D E 9. A B C D E 20. A B C D E 31. A B C D E 10. A B C D E 21. A B C D E 32. A B C D E 11. A B C D E 22. A B C D E Student-Produced Responses 1.2 / 5 / © Pearson Education, Inc., publishing as Pearson Prentice Hall. / . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 4 4 4 3. 5 / . . 2. / . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . . 7 / 1 0 8 / / . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . . 4. 3 / / . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . . 5. / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 3 4 4 44 4 . . 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 6. 7 2 0 / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 4 4 4 . 7. 2 0 / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . 8. . 7 8/ / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . 9. 64 / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . 10. 6 / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 3 4 4 4 4 4 . 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 99 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 Answers Geometry 163 Name Class Date © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Answer Sheet 1. A B C D 27. A B C D 2. F G H J 28. F G H J 3. A B C D 29. A B C D 4. F G H J 30. F G H J 5. A B C D 31. A B C D 6. F G H J32. F G H J 7. A B C D 33. A B C D 8. F G H J 34. F G H J 9. A B C D 35. A B C D 10. F G H J 36. F G H J 11. A B C D 37. A B C D 12. F G H J 38. F G H J 13. A B C D 39. A B C D 14. F G H J 40. F G H J 15. A B C D 41. A B C D 16. F G H J 42. F G H J 17. A B C D 43. A B C D 18. F G H J 44. F G HJ 19. A B C D 45. A B C D 20. F G H J 46. F G H J 21. A B C D 47. A B C D 22. F G H J 48. F G H J 23. A B C D 49. A B C D 24. F G H J 50. F G H J 25. A B C D 51. A B C D 26. F G H J 52. F G H J Answer Sheet Geometry 165 Blank Sheet of Grids 2. 4. / . / . / . . . / . / . / . . . / . / . / . . . / . / . / . . 0 1 2 3 45 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 23 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 6. 7. All rights reserved. . 5. 8. . / . / . / . . . / . / . / . . . / . / . / . . . / . / . / . . 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 45 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 9. 166 3. 10. 11. 12. . / . / . / . . . / . / . / . . . / . / . / . . . / . / . / . . 0 1 2 3 4 5 6 7 8 9 0 12 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Geometry Answer Sheet © Pearson Education, Inc., publishing as Pearson Prentice Hall. 1. Student Answer Sheet: SAT/ACT Practice Test All rights reserved. Multiple Choice 1. A B C D E 12. A B C D E 23. A B C D E 2. A B C D E 13. A B C D E24. A B C D E 3. A B C D E 14. A B C D E 25. A B C D E 4. A B C D E 15. A B C D E 26. A B C D E 5. A B C D E 16. A B C D E 27. A B C D E 6. A B C D E 17. A B C D E 28. A B C D E 7. A B C D E 18. A B C D E 29. A B C D E 8. A B C D E 19. A B C D E 30. A B C D E 9. A B C D E 20. A B C D E 31.A B C D E 10. A B C D E 21. A B C D E 32. A B C D E 11. A B C D E 22. A B C D E Student-Produced Responses 1. 2. / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 4 4 4 © Pearson Education, Inc., publishing as Pearson Prentice Hall. . 3. / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . 4. / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 33 3 4 4 4 4 . 5. / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 3 4 4 4 4 4 . 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 / / . . .. 0 0 0 6. 7. / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 4 4 4 . 8. / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . 9. / / . . . 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 . 10. / / . . . 0 0 0 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 . 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 67 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 Answer Sheet Geometry 167 Answers Activity 1 1. a. 12 ft; 2 s b. 8 ft; 3 s 2. Check student’s work. 3. Walk at a constant speed away from the motion detector. 4. Check student’s work. 5.Divide the change in distance by the change in time; ft/s 6. Check student’s work. 7. The speed equals the slope of the line; vertical change slope = . horizontal change 8. a line with a steeper slope 9. Students should sketch a line with a steeper slope 10. Students should sketch a line with a less steepslope. 11–12. Check student’s work. 13. Walk toward the motion detector. 14. Stand still; impossible, because the person would have to be in two places at the same time. 15. the initial distance from the motion detector 16a. stood still, then walked at a constant speed away from the detector, then stoodstill again 16b. walked away from the detector at a constant speed, stood still, walked toward the detector at a faster, constant speed 16c. walked away from the detector at a constant speed, stood still, walked away from the detector at a much faster, constant speed 17. Yes; the distance at the beginningand end of the graph is the same. 18. Toward the detector; the steepness is greater so the speed is greater. Take a Walk, Part 2 1. 2. 3. 4. 5. 6–8. 9–10. 11. 12–14. 15. 16. 104 Activity 2 negative; positive the line through A; 6 yes; y = 6x Check students’ work. Check students’ work; initial distance from themotion detector Check students’ work. Check students’ work. If an equation models the data well, the line should overlap the plot. If an equation does not model the data well, the line may either have a different slope or an incorrect y–intercept. The model will predict reasonable values for x 0 and x thefinal time measured by the CBL 2 unit. Other values of x predict values before or after the student was walking. Check students’ work. No; the y–intercept will be negative and does not represent the initial distance from the motion detector to the student, because the student stood still for 2 s. Yes; theslope is the speed of the student after he or she began walking. Answers Synchronized Strut Activity 3 1. Lines have different positive slopes and have the same y–intercept. 2. parallel with different y-intercepts 3. Same line 4–5. Check students’ work. 6. y–intercept; represents initial distance from themotion detector 7–9. Check students’ work. 10. Speed each student walks 11. They walk at different speeds. 12. Check students’ work. 13. the student whose graph has a steeper slope 14–15. Check students’ work. 16. The slopes should be about the same because they walked at the same speed. 17.They have the same slope. 18. m 19. The difference is constant because they are walking at the same speed. 20. They are the same line. Coming and Going Activity 4 1. No, a motion detector is like a function machine because there is only one distance for any given time. 2. Students should sketch a lineof positive slope with a y–intercept near zero. They should also sketch a line of negative slope with a y-intercept near the top of the screen. The negative slope indicates a person walking toward the detector. The positive slope indicates a person walking away from the detector. The y–intercepts show howfar from the detectors each person started. 3. Not necessarily. It can be approximated from the data, however. 4–7. Check students’ work. 8. Check students’ work. If an equation models the data well, the line should overlap the plot. If an equation does not model the data well, the line may either have adifferent slope or an incorrect y–intercept. 9–10. Check students’ work. 11. The measurements should agree closely with the calculations. Discrepancies might be caused by delays due to the reaction times of the marker and timer. 12. A motion detector records only the distance to the closest object infront of it. A motion detector cannot record two objects at once. A function has only one y–value for each x–value. 13. Check students’ work. Keep Your Eye on the Ball Activity 5 1. The student should sketch a series of parabolas opening downward with maximum values that gradually decrease. 2.decrease 3. Yes, each rebound is shorter than the previous bounce. 4–5. Check students’ work. Technology Activities © Pearson Education, Inc. All rights reserved. Take a Walk, Part 1 Answers (continued) 6. vary 7. decreasing 8. Multiply a height value by 75% or 80% and see if the result is the height ofthe previous bounce. 9. yes; constant 10. Check students’ work. 11. b = -2ax; check students’ work. 12–16. Check students’ work. © Pearson Education, Inc. All rights reserved. Race Cars Activity 6 1. Yes; the battery-powered car gives a linear graph because it travels at a constant speed. The speed ofthe wind-up car changes, so its graph will show a curve as it speeds up and a curve as it slows down. 2. Students should sketch a line for the battery-powered car and a curve showing the acceleration and deceleration of the wind-up car. 3. The battery-powered car can be modeled by a line. The wind-upcar can be modeled by two parabolas, one for acceleration and one for deceleration. 4. Check students’ work; a line 5. Check students’ work; the slope is the speed and the y–intercept is the initial distance from the motion detector. 6. Check students’ work. The equations may not model the data after thecar initially starts driving because there is some acceleration. 7. Check students’ work. 8. The values should be very close. The slope might be higher than the average speed from the graph because the slope ignores the brief acceleration of the car. 9. 0 x 1.5; students may want the domain to extendbeyond x = 1.5 because the car continues to drive. Any domain that they can justify is reasonable, but it should not be very long, since the car may stop. 10. The collected data would span a longer period of time. Everything else should remain constant. 11. Check students’ work. Should be approximatelythe same as the previous model. 12. Check students’ work; yes. 13. 0 x 6 14. Check students’ work. 15. The data will appear linear but you would expect some type of curve. You will see a line because you are recording data for a very short time. But the springloaded car must be either speeding up orslowing down since it does not go at a constant speed as does the battery-powered car. Thus you would expect to see a curve. 16. A longer time interval will give a better look at the car’s motion. You will probably see s-shaped data. You will see a parabola opening upward for when the car starts and isspeeding up and a parabola opening downward for when it is slowing at the end. Technology Activities Back and Forth It Goes Activity 7 1. Sketch should resemble a cosine curve with negative amplitude. 2. decrease; increase 3. Check students’ work. 4. Students should label the minimums andmaximums of the graph. The motion detector records points at regular intervals. It probably does not record the actual minimum and maximum but records points near the maximum and minimum. 5–9. Check students’ work. 10. Check students’ work; yes; non-linear 2 11. y = g a x b ; check students’work; it should be a 2p good model. 12. 1.74; the two methods should give close answers but there will be some experimental discrepancy. 13. l = 0.64; the answer is reasonable, although it does not agree exactly with the measurement from the activity. l ; independent; as you change the length of Äg thependulum, the period changes. 15. Check students’ work. The model gives T = 2.84 s. The model and the experiment should give similar answers. 16. You should shorten it. 14. t = 2p Charge It! Activity 8 1. f(x); g(x) 2. a is the starting amount or the y–intercept. b is the base, the growth factor, or the decayfactor. 3. b 1; 0 b 1 4. 1; 2; y = 2x 5. 30; 0.75; y = 30(0.75)x 6–7. Check students’ work. 8. Check students’ work; it is the voltage of the battery. 9. The two answers should be close. The y–intercept may be lower because the capacitor may still have been collecting charge. 10. They are the same. 11.exponential decay 12–15. Check students’ work. 16. They should be very close. 17. No, exponential functions never reach their asymptotes. No, the voltage of the capacitor will eventually reach zero. 18. 0.9782; they should be close. Falling Objects Activity 9 1. Students should graph a horizontal line witha positive y–intercept that changes to the right side of a downward-opening parabola. There should be a horizontal line once the parabola reaches the x–axis. Answers 105 (continued) 2. quadratic 3. Check students’ work. First the book was held over the motion detector. Next it fell. Finally it lies on top ofthe motion detector. 4–11. Check students’ work. 12. The vertex represents the time at which the student dropped the book and the book’s initial height above the motion detector. 13. Check students’ work. Only the part of the parabola that corresponds to the actual time the book was falling should beincluded in the domain. At the other times, the book was actually stationary. 14. Check students’ work. 15. The model should not agree exactly. The physics model assumes that there is no air resistance. The book is affected by air resistance. The physics model assumes that the book starts to fall when t =0. In the activity, the book was held briefly so that it started to fall after t = 0. Bouncing Ball, Part 1 Activity 10 1. The student should sketch a series of parabolas that open upward that all stop about 5 units above the x–axis. 2. The student should sketch a series of parabolas opening downward with

maximum values that gradually decrease. Students who understand that the motion detector measures distance from itself should understand that the final result will be a graph of the distance from the floor instead of the distance from the motion detector. 3. The ball will bounce to a maximum height thatis a percent of the previous height. The percent should be constant. 4–5. Check students’ work. 6. One method is to substitute the coordinates of the vertex for h and k; next substitute the coordinates of a point on the parabola for x and y; then solve for a. Another method is to find three points on theparabola and substitute their coordinates into the equation y = ax2 + bx + c; then solve for a, b, and c. 7. Check students’ work. The curve is too wide and opens up instead of down. 8. A parabola with a negative a opens down. A parabola with a positive a opens up; negative 9. The curve is narrower. Alarger a means a narrower curve. A smaller a means a wider curve. 10. Check students’ work, but a should be near -16. 11. No; if students use an actual data point, their models may be off in height or not centered on the bounce horizontally. 12. Check students’ work. They should be similar. 13. a is theacceleration due to gravity, h is the time when the ball is at the maximum height, k is the maximum height of the ball for the bounce. a should remain constant because gravity is constant. 106 Answers Bouncing Ball, Part 2 Activity 11 1. The student should sketch a series of parabolas opening downwardwith maximum values that gradually decrease. 2. eventually comes to a stop 3. decrease 4. remains constant 5–8. Check students’ work. 9. exponential or possibly quadratic 10. A quadratic equation does not have an asymptote. After an infinite amount of time, a quadratic equation predicts that the ball isan infinite distance below the floor. After the same amount of time, an exponential equation predicts that the ball is at 0 ft, or resting on the floor. 11–13. Check students’ work. 14. Check students’ work. A well-fitting model should approximately touch the vertex of each bounce. 15–16. Check students’work. 17. coefficient of restitution When’s the Tea Ready? Activity 12 1. 100°C; room temperature. A liquid cannot cool below room temperature. 2. Students should draw an exponential function with an asymptote at room temperature. 3. The graphs will be similar, but the probe should cool more quicklybecause it has less mass. 4–5. Check students’ work. 6. exponential 7. the x–axis 8. Check students’ work; no. 9. By adding a constant c, the graph of the exponential equation is translated vertically and the asymptote is no longer the x–axis. 10. c should be given the value found in Exercise 8; it is thetemperature of the classroom. 11–13. Check students’ work. 14. Check students’ work. The model should fit well because it overlaps most of the data points. The model may not fit the first few points. 15. Check students’ work. 16. No, a is the difference between the initial value and the room temperature,c. 17. Check students’ work. This model is a vertical translation of the previous model. The translation is necessary because the room temperature is greater than zero. 18. yes; yes 19. yes; yes 20–21. Check students’ work. 22. a is the difference between room temperature and the initial temperature ofthe water. b is the rate at which the water cools; it does not change, c is room temperature; it does not change. Check students’ work. Technology Activities © Pearson Education, Inc. All rights reserved. Answers Answers (continued) Full Speed Ahead 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. © PearsonEducation, Inc. All rights reserved. 14–15. 16. Activity 13 Students’ should sketch a line. Students should sketch a curve opening upward. Students should sketch a curve opening downward. Check students’ work; yes. Check students’ work; the slope is the speed and the y–intercept is the initial distancefrom the motion detector. The model should fit reasonably well, the residual plot should show a definite parabolic pattern. quadratic because the speed is increasing Check students’ work. The model should fit better. The residuals should confirm the quality of the model. The speed is increasing because asx increases the change in y divided by the change in x is getting larger. This means that the car is traveling farther during each x-unit. The sketch should show a parabola opening upward representing an increase in speed joined to a parabola opening downward. The data shows the car speeding up at thestart and then slowing down. The model is reasonable only over the domain 0 x 1.5. The time interval was too short to see the car slow down. Check students’ work. Students should label when the car is accelerating and decelerating. Check students’ work. Models should fit well. Students can use theintersect option on the graphing calculator or solve the system of quadratic equations formed by the two models. Check students’ work. In the Swing of Things Activity 14 1. The graph should be a periodic function with a maximum or minimum at the y–axis. 2. the length of the string 3. cosine or sine 4. a =amplitude; b = 2p divided by the period; c = horizontal translation; d = vertical translation 5–7. Check students’ work. 8. Take half the difference between the y–value of a peak and the y–value of a trough. Check students’ work. 9. Take the difference between the x–values of two peaks or two trough.Check students’ work. 10. Find the x–value of the first peak. Check students’ work. 11. Take the average of the y–values of one peak and one trough. Check students’ work. 12–13. Check students’ work. 14. The amplitude is greater, because the pendulum is pulled back farther. The graph has a greatervertical translation because the motion detector sill needs to be 1.5 m away from the closest swing of the pendulum. 15. The period decreases because the string is longer. The amplitude and vertical translation increase because a longer string means that the pendulum starts farther away from the centerof its swing. Technology Activities 16. The magnitude of a is determined by the angle at which the string is pulled back and by the length of the string. b decreases with the length of the string. c increases when the motion detector is started after the pendulum is swinging. d increases when the string ispulled back farther and when the string is longer. Playing With Numbers Activity 15 Investigate 1. See screen 2. The digits are the same but in a different order. 3. The digits are the same but in a different order. Exercises 1–3 4. Yes 5. Same digits with zero included Exercise 4 Exercise 5 6. Multiples of 3have been omitted from the list. 7. 22A, 23A, 25A, 26A 8. The same patterns occur. Investigate 1. See screen Exercise 1 Exercise 4 2. 111112 = 123454321 3. The paper and pencil calculation of 1112 is as follows: 111 3111 111 111 111 12321 Notice that the number of one’s in each column increases(then decreases) by 1 from right to left. 4. It can be seen that the digits increase (then decrease) by 1 from left to right. 5. The last 7 in the number has been rounded up from 6. Answers 107 (continued) 6. Using paper and pencil, it can be seen that there is a column of 10 one’s, whose sum is 10. Thisresults in a “carry-over” which causes the pattern to break down. Investigate 1. 2. 3. 4. Answers may vary. Sample: 82471 Answers may vary. Sample: 27814 Answers may vary. Sample: 82471 - 27814 = 54657 Answers may vary. Sample: 5 + 4 + 6 + 5 + 7 = 27; 72 - 27 = 45; 4+5=9 5. A sum of 9 alwaysresults. 6. Variables can take on any numerical value, therefore proving a result for any number. Exercises 1. The result contains five copies of the two-digit number (i.e. 48 ? 101010101 = 4848484848). 2. The number is repeated as a decimal (i.e. 4 = 4). 9 They do not terminate; they keep repeating. 3.Yes. In this case 9 = 1 and .9 = 1 as well. 9 4. The number is repeated as a decimal (i.e. 36 = .36). 99 5. Yes. In this case 99 = 1 and .99 = 1 as well. 99 Finding Real Roots of Equations Activity 16 Investigate 1–3. Check student’s work. 4. .8241323123 5. -.8241323123 Investigate Investigate 1. Checkstudent’s work. 2. .8241323123 3. -.8241323123 Exercises 1. They are approximations. 2. .3862368706; 1.961569035 3. The (x, y) coordinates of the graphs are real numbers, so graphical methods can only reveal real answers. Activity 17 Investigate 1–3. Check student’s work. Investigate 4. Checkstudent’s work. 5. All values of x where the graph is below the x-axis. 108 Answers Investigate 7. Check student’s work. 8. The inequality returns a value of -1 for 1 x 3 and 0 everywhere else. The points on the graph with y-coordinate 0 are hidden by the x-axis. Exercises 1. -2 x 1; methods of solution mayvary 2. - 17 x 57 3. 43 x 3 4. 1 x 2 Magic Pricing Numbers Activity 18 Investigate 1. 1 or 24–1 24 2. 1.16 3. .9 4. 1.028 5. 1 ? 1.16 ? .9 ? 1.028 = .044718 24 6. y = .044718x 7. Here are two examples to describe the rounding process. The number 31.243 typically rounds down to 31.24 (to the nearestpenny). By adding .005, you get 31.243 + .005 = 31.248 which rounds up to 31.25. The number 31.247 typically rounds up to 31.25. By adding .005, you get 31.247 + .005 = 31.252 which still rounds to 31.25. 8. Yes 9. $28.81 Exercises 1. $21.62 1 ? 1.16 ? .9 ? 1.028 = .029812; 2. Multiply 857.98 by 36 y= .029812(857.98) = $25.58 3. w = (.044718)–1 = 22.36236x, where w = wholesale price and x = retail price 6. Check student’s work. 7. -.8241323123 Solving Absolute Value Inequalities Graphically 6. All values of x where the graph is on or above the x-axis. Mrs. Murphy’s Algebra Test Scores Activity 19Investigate 1. first line: 30 units; second line: 20 units 2. x - 42 42 3. x 2 30 y 2 78 20 y 2 78 x 2 42 5. 30 = 20 6. The graph is a straight line. 4. y - 78; Exercises 1. 60 Technology Activities © Pearson Education, Inc. All rights reserved. Answers Answers (continued) y 2 78 2. x 2 32 = or y = 1 (x - 32) + 78 240 20 4–6. Check student’s work. Exercises 1. y 2 72 3 or y = (x - 40) + 72 3. x 2 40 = 36 4 27 Exploring Point-Slope Form Activity 20 Investigate 1. y - 1 = 3(x - 2) 2. 2–4. Check student’s work. 5. L3 = {-2, -1, -.5, 0, .5, 1, 2} Exercises 1. 3. 2. 4. 3. Burgers cost $2.95; Fries cost $1.80; Soft Drinks cost $1.105. If it costs $25.04 for 4 burgers, 4 fries, and 4 soft drinks, then it should cost 25.04 2 = $12.52 for 2 burgers, 2 fries, and 2 soft drinks. The screen at right shows the error message that results when [A]–1[C] is executed. © Pearson Education, Inc. All rights reserved. 4. Wheat on a Chessboard Activity 22Investigate 1. 5. Use L3 = {-2, -.5, 0, .5, 2, 200} and Y1 = L3 * (X - 12) + 12 Fast Food Follies Activity 21 Investigate 1. 3x + 4y + 4z = 16.45 5x + 2y + 5z = 19.30 4x + 6y + 6z = 23.50 2. The product of matrix A and matrix X will produce matrix B. 3. If AX = B, then X = A–1B. Technology Activities Square 12 3 4 5 6 7 8 9 10 Num. of Grains 1 2 4 8 16 32 64 128 256 512 2. Num. of Total Num. Squares of Grains 1 1 2 3 3 7 4 15 5 31 6 63 7 127 8 255 9 511 10 1023 Answers 109 (continued) 3. Raise 2 to the number of squares and subtract 1. 4. T(n) = 2n - 1 where n = number of squares and T(n) = totalnumber of grains; about 1.8 1019 grains of wheat 5. about 1.7 1013 or 17 trillion bushels 6. 6728.3 years Exercises 1. 2. 3. 4. 5. 6. 1,075,000 21 20 255 4,294,967,295 (which is 232 - 1) The wheat covering 51 squares provides a close match to the 1998 U.S. wheat production. 7. about 2.6 inchesManipulating a Polynomial Activity 23 Investigate 1. Check student’s work. 2. It moves up. 3. Evaluate f(0) = 12 to find the y-intercept, (0, 12), and 12 10. 4. The y-values range from -20 to 20 and the scale is 1, which results in 40 tick marks on the y-axis. 5. Check student’s work. Investigate 1. Because thepolynomial has a value of 0 when x = -1. 2. Check student’s work. 3–4. Answers may vary. Sample: f(x) = (5x4 - 17x3 + 13x2 - 8x + 12)(x + 1)(x + 3); graph shown with Xmin = -4.7, Xmax = 4.7, Xscl = 1, Ymin = -440, Ymax = 100, Yscl = 0 3. Writing a Simple Program Activity 24 Investigate 1–8. Checkstudent’s work. Exercises 1. Answers may vary. Sample: A = 1, B = 1, and C = -20 has roots -5 and 4. 2. Answers may vary. Sample: A = 4, B = -12, and C = 9 has root 3 . 2 3. Answers may vary. Sample: A = 2, B = -5, and C = 7 has imaginary roots 5 - 1.392i and 5 + 1.392i. 4 4 4. Answers may vary.Sample: A = 6, B = 7, and C = -5 has rational roots - 5 and 1 . 3 2 5. Answers may vary. Sample: A = 3, B = 8, and 24 4 !37 C = -7 has irrational roots 3 (-3.361 and .694). 6. a. $625.72 b. $2236.58 Repeated Radicals Activity 25 Exercises 1. 9 2. Exercises 1. Answers may vary. Sample: f(x) = (x + 3)(x - 1)(x - 3)(x - 5); graph shown with Xmin = -4.7, Xmax = 6, Xscl = 1, Ymin = -120, Ymax = 20, Yscl = 0 2. No portion of the graph can be seen in the ZDecimal window. 110 Answers x = 6 + $6 1 #6 1 "6 1 !6 1 % (x - 6)2 = a $6 1 #6 1 "6 1 !6 1 %b 2 x2 - 12x + 36 = 6 + #6 1 "6 1 !6 1 % x2 - 12x + 36 = x x2 - 13x+ 36 = 0 (x - 4)(x - 9) = 0 x = 4 and x = 9 The solution x = 4 is extraneous, since x is clearly greater than 6. 3. 5.302775638; the number is an approximation 7 1 !13 4. ; irrational 2 5. The calculator can only give an approximate answer, whereas algebraic techniques give an exact answer. TechnologyActivities © Pearson Education, Inc. All rights reserved. Answers Answers (continued) Asymptotes and Holes Activity 26 Investigate 1. The function is undefined at x = 42. This makes sense algebraically since the denominator is equal to zero at these two values of x. 2. It disappears. 3. The functionapproaches 4 on either side of the singularity. This makes sense because the denominator approaches zero as x approaches the singularity. 4. Vertical lines appear at the singularities. 5. The y–value ‘jumps’ from negative to positive (or vice versa) as you trace through a singularity. 6. The calculator wasdrawing a line between two points on either side of the singularity. 7. Yes; the singularity at x = -2 appears as a hole in the graph. 8. The function is undefined when the denominator is equal to zero. 9. As you approach the singularity at x = -2 from the left or right, the corresponding y–values approach thesame number. As you approach the singularity at x = 2 from the left, the y–values approach -. As you approach the singularity at x = 2 from the right, the y–values approach +. 10. No; the simplified function is defined at x = -2 and the hole disappears. 11. When graphing Y1, the graphing ball ‘breaks’ whenit reaches x = -2. This does not happen when you graph Y2. Exercises © Pearson Education, Inc. All rights reserved. 1. The first graph has a hole at x = 1 whereas the second x2 2 1 is all real graph does not. The domain of y = x21 numbers except 1 and the domain of y = x + 1 is all real numbers. 2 2.Answers may vary. Sample: f(x) = (x 1 1) (x 2 1) 3 2 3. Answers may vary. Sample: f(x) = x 2 x or x 2 1 x 2 (x 2 1) f(x) = x21 4. Check student’s work. 5. Type ‘.05 S 䉭X’ on the home screen to eliminate false asymptotes. Stepping Out Activity 27 Investigate 1. The graph resembles a series of steps. 2. Therange values are {1, 2, 4, 3, 4}. Answers may vary. Sample: ‘Int’ rounds numbers down to the nearest integer. 3. The range values are {-2, -2, -5, -4, -5}. Answers may vary. Sample: ‘Int’ rounds numbers to the nearest integer that is less than or equal to the number. Technology Activities 4. The calculatorwill connect the endpoints of each step to the next step. 5. 6. $6; $11; the function C(x) = 3.50 + 2.50x models a cost that varies continuously as a function of mileage, whereas the actual cost increases in steps. 7. The function rounds the mileage down, whereas the cab company rounds the mileage up.For example, a 4.4 mile trip is charged at the same rate as a 4 mile trip instead of a 5 mile trip. 8. C(x) = 3.50 + 2.50int(x + 1) Exercises 1. C(x) = .34 + .23int(x + 1) 2. The two functions are identical except at the integer values. A careful graph of y = int(x + 1) would show a closed dot on the left of each stepand an open dot on the right, while a careful graph of y = -int(-x) would show an open dot on the left of each step and a closed dot on the right. 3. Answers may vary. Sample: Y1 = 3.5 - 2.5int(-x); Answers may vary. Sample: Y1 = .34 - .23int(-x) Systems With Many Solutions Activity 28 Investigate 1. Theerror message ‘ERR:SINGULAR MAT’ appears. 2. Yes 3. 2x + 3y + z = 6.85 4x + 3y + 3z = 11.65 3x + 3y + 2z = 9.25; Yes; the prices given in the previous question solve the system. 4. Because matrix A does not have an inverse. 5. $6.85; $11.65; $9.25 6. $6.85; $11.65; $9.25 7. There can only be onematrix [A]–1[B], so the calculator could not produce multiple correct answers in this way. Answers 111 (continued) Exercises 1. Check students’ work. 2. Any values of x, y, and z satisfy the equation. 3. 3 ay 2 1 zb = 3 a 41 b S 3y - z = 41 S 3 20 60 3y - z = 2.05 4. x + z = 12 5 5. x = 2.2; z = .2 6. 2(2.2) +3(.75) + 2 = 6.85 S 6.85 = 6.85 4(2.2) + 3(.75) + 3(2) = 11.65 S 11.65 = 11.65 3(2.2) + 3(.75) + 2(2) = 9.25 S 9.25 = 9.25 Extend 7. the cost of a burger is too low; some of the numbers are not rounded to the nearest penny. 8. Answers may vary. Sample: $1.30, $1.05, $1.10; $1.60, $0.95, $0.80; $1.90,$0.85, $0.50 Parabolic Reflectors Activity 29 Investigate tional. This explains why no graph results. 6. The calculator only evaluates y = (-1)x for values of x that only have real answers (such as x = 2 , 3 , 4 , c). 5 5 5 These values cause y to alternate between -1 and 1. 7. The calculator only evaluates y =(-1)x for values of x whose fractions reduce such that the numerators are always even numbers (such as x = 2 , 4 , 6 , c). These 5 5 5 x–values always produce y–values of 1. 8. The calculator only evaluates y = (-1)x for values of x whose fractions reduce such that the numerators are always odd numbers(such as x = 1 , 3 , 5 , c). These 5 5 5 x–values always produce y–values of -1. Exercises 1. y = (-2)x; y = (-2)–x; y = 2(-1)x 2. p is irrational—it cannot be written as the quotient of two integers; use a rational approximation for p such as 314159 . 100000 Bull’s-eyes 1–4. Check students’ work. 5. They areequal. 6. Check students’ work. Activity 31 Investigate Exercises 1. (0, 4) 2. closer 3. By placing the lamp at the focus of a parabolic reflector, all rays that strike the reflector are directed outward parallel to the axis. This helps ‘concentrate’ the light in a specific direction. Exploring Powers of –1 1. Checkstudents’ work. 2. Check students’ work. 3. Activity 30 Investigate 1. Check students’ work. 2 1 2. (21) 6 = -1; f (21) 2 g 6 = 1; The calculator 2 1 evaluates (21) 6 as (21) 3 = -1, whereas the law of exponents would seem to imply that 2 1 1 (21) 6 = f (21) 2 g 6 = 16 = 1. The answer depends on the fractionbeing reduced to lowest terms before the laws of exponents are applied. 3. It reduces 2 to 1 , which is the same as taking the cube 6 3 root. 4. Values of x are considered as fractions in lowest terms before being used as exponents. For example, f(.4) is 4. the second and third one 5. type the functionshown below evaluated as f a 2 b = (21) 5 = 1. f(.5) is evaluated 5 2 as f a 1 b = (21) 2 which is undefined and f(.6) is 2 1 evaluated as f a 3 b = (21) 5 = -1. 5 5. Using this window, the calculator attempts (but fails) to evaluate y = (-1)x for values of x that are irra3 112 Answers Technology Activities ©Pearson Education, Inc. All rights reserved. Answers Answers (continued) Exercises Exercises 1. D: -2 x 2 R: 0 y 2 D: -4 x 4 R: 0 y 4 D: -4 x 4 R: 0 y 2 D: -2 x 2 R: 0 y 4 (x 2 2 1) (x 2 2 4) x6 1 1 2. The graph crosses its own asymptote an infinite number of times. 3. Yes, because the function getsarbitrarily close to y = 2 as x approaches 4. The fact that the function actually equals 2 is technically not an issue. 4. Answers may vary. Sample: x2 if x 0 1. Answers may vary. Sample: y = Making Ellipses out of Circles Activity 34 Investigate 1–3. Check students’ work. 4. It looks identical to the graphshown in the introduction. 5. A horizontal ellipse with major axis of length 6 and minor axis of length 4 is a circle of radius 1 that is stretched horizontally by a factor of 3 and vertically by a factor of 2. These scale factors can be produced by dividing the x–axis window values by 3 and the y–axis windowvalues by 2. 2. Exercises 1. Divide Xmin, Xmax, and Xscl by 2; divide Ymin, Ymax, and Yscl by 3. 3. Check students’ work. The Natural Logarithm Activity 32 © Pearson Education, Inc. All rights reserved. Investigate 1. 2. 3. 4. Check students’ work. 1 Check students’ work. Upper limit = e 2. Divide Xmin,Xmax, and Xscl by 4; divide Ymin, Ymax, and Yscl by 3. Exercises 1. .6931471806; ln(2) = .6931471806; the two numbers are equal 2. they are equal 3. they are equal 4. it is negative 5. For any a 0, ln(a) = the area under the graph of 1 between x = 1 and x = a, with “area” being y= x negative for 0 a 1.Can a Graph Cross Its Own Asymptote? 3. Divide the X–Window values by 6 and the Y–Window values by 2. The graph does not fit in the window. Activity 33 Investigate 1. The x–axis 2. no 3. yes; Y2 = 4. twice 5. three times 6. yes x (x 2 1 1) Technology Activities Answers 113 Answers (continued)Exercises 4. First, multiply Xmin, Xmax, Ymin, and Ymax by 2. Then follow the steps in the answer to Exercise #3 above. 1. 2. 3. 4. Check students’ work. 5.187377518; ln(100) 艐 4.605, 1 + ln(100) 艐 5.605 6.79282343; ln(500) 艐 6.215, 1 + ln(500) 艐 7.215 No; it’s between ln(109) = 20.723 and 1 +ln(109) = 21.723. 5. e100 艐 2.688 1043 terms Monte Carlo π 5. First, multiply Xmin, Xmax, Ymin, and Ymax by 2. Then divide Xmin, Xmax, and Xscl by 2 and divide Ymin, Ymax, and Yscl by 6. 6. First, multiply Xmin, Xmax, Ymin, and Ymax by 2. Then divide Xmin, Xmax, and Xscl by 2 and divide Ymin,Ymax, and Yscl by 6. Finally, add 3/6 to Ymin and Ymax to re-center the screen vertically. The circle command will be ‘Circle(2/2, 3/6, 1).’ The Harmonic Series Activity 35 Investigate 1. The graph rises (slowly) without bound. 2. y = ex; all reals 3. 1, 1 , 1 , c ; 1, 1 , 1 , c 2 3 2 3 4. In both figures, the area ofthe rectangles represents the terms of the harmonic sequence. The expression ln(a) represents the area under the curve from x = 1 to x = a. Since the rectangles are all above the curve, 1 ln(a) must be the inequality 1 1 1 1 1 1 % 1 a 2 3 true. In the second inequality, since the rectangles are entirelycontained within the region below the curve, 1 1 ln(a), the inequality, 1 1 1 1 1 1 % 1 a 2 3 must be true. 5. between 9.21 and 10.21 114 Answers Activity 36 Investigate 1. The probability that a randomly selected point lies within the shaded region is equal to the ratio of the area of the shaded region to thearea of the square, p that is, 4 5 p 4 1 2. Check students’ work. 3. Because the coordinates, A and B, will always be less than or equal to 1. 4. Answers may vary. Sample: The ratio will most likely fall between 2 and 5, with most results clustered near 3.2. 5. Answers may vary. Sample: The ratio shouldconsistently be very close to 3.14. Exercises 1. Line 1: Clears any drawings Line 2: Turns off all plots Line 3: Turns off any functions located in the ‘Y=’ editor Line 4: Sets the viewing window, graphs square, graphs circle Line 5: Set the value of H equal to 0 Line 6: Prompts the user to input the number ofpoints for the simulation Line 7: Sets up a loop that will execute the next 4 lines N times; K is a counter Line 8: Stores random numbers between 0 and 1 as A and B Line 9: Plots the ordered pairs (A, B) on the graph Line 10: Calculates the distance from the origin to point (A, B) and tests if this distance isless than or equal to 1 (determines if the point is within the circle) Line 11: Adds 1 to the value of H if the point falls within the circle. Line 12: Specifies where the loop ends, which happens when K is equal to N Lines 13 and 14: Displays “4 * RATIO IS:” along with the value of 4H/N (which should beapproximately equal to p) 2. Check students’ work. The Way the Ball Bounces Activity 37 Investigate 1–2. Check students’ work. 3. A ball bouncing on a spring. Technology Activities © Pearson Education, Inc. All rights reserved. Extend Answers (continued) 4. The ball moves quicker as it passes throughthe origin because its motion is mostly in the vertical direction (keeping in mind that that ball is really traveling along a unit circle). The ball moves slower at the top and bottom of its path because its motion is mostly in the horizontal direction. 5. Yes 1. 2. 3. 4. The ball moves twice as fast. Change Tmax to18p. 3 cycles of a sine curve. Nothing; the scale along the x–axis is so large that the graph is entirely contained along the y–axis and cannot be seen. 5. The wave has been compressed horizontally (because of the large x–axis scale), giving a result identical to step 3 of the investigation. 6. The verticalposition of a point traveling along a unit circle at a constant speed can be used to model harmonic motion. If the vertical motion is graphed as a function of time, it generates a sine wave. Graphing Ellipses and Hyperbolas Activity 38 Investigate © Pearson Education, Inc. All rights reserved. 1–2. Checkstudents’ work. 2 (3cosT) 2 (2sinT) 2 y2 5 1 3. x 1 = 9 4 9 4 2 2 9cos T 1 4sin T 5 cos2T + sin2T = 1; horizontal 9 4 stretch factor = a, vertical stretch factor = b 1 2 sin2T 5 1 2 sin2T 5 cos2T cos2T cos2T cos2T 5 1; a hyperbola cos2T 5. Check students’ work. 4. x2 - y2 = Exercises 3 ; Y1T = 2sinT 5 2tanTcosT cosT Vertical Angles They are congruent. m⬔EAD + m⬔EAB = 180 m⬔EAD + m⬔BAC = 180 (⬔EAD + ⬔EAB) - (⬔EAB + ⬔BAC) = 0 (⬔EAD + ⬔EAB) - (⬔EAB + ⬔BAC) = 0 simplifies to ⬔EAD - ⬔BAC = 0 which means ⬔EAD = ⬔BAC Technology Activities Activity 40 Investigate 1. Themeasure of an exterior angle of a triangle is equal to the sum of the remote interior angles. 2. supplementary; ⬔A + ⬔B + ⬔ACB = 180 and ⬔ACB + ⬔BCD = 180; rewrite both equations as ⬔A + ⬔B = 180 - ⬔ACB and ⬔BCD = 180 - ⬔ACB; by substitution ⬔A + ⬔B = ⬔BCD Extend 3. their sum is 90;they must be complementary since the angles of a triangle have a sum of 90 and ⬔B = 90 4. their sum is 270; ⬔DAC + ⬔ACE = (⬔BAC + ⬔B) + (⬔ACB + ⬔B) = (⬔BAC + ⬔ACB) + (⬔B + ⬔B) = 90 + 180 = 270 Investigate Further 5. ⬔ADE - (⬔A + ⬔B + ⬔C) = 180; ⬔ADE + 180 = ⬔A + ⬔B + ⬔C;⬔A + ⬔B + ⬔C + ⬔ADC = 360 and ⬔ADC + ⬔ADE = 180; ⬔A + ⬔B + ⬔C = 360 - ⬔ADC and ⬔ADE = 180 - ⬔ADC; by substitution ⬔ADE + 180 = ⬔A + ⬔B + ⬔C 6. the sum of the remote interior angles of a pentagon is equal to the sum of the exterior angle and 360. 7. Answers may vary. Sample:No. of Sides Ext. ⬔ Sum of Remote Int. > 3 4 5 6 80 80 80 80 80 260 440 620 Ext. ⬔ Sum of Remote Int. > 0 180 360 540 Activity 39 Investigate 1. 2. 3. 4. 5. 6. Check students’ work. 7. They are perpendicular. 8. Answers may vary. Sample: Adjacent angles are supplementary. The angle bisectors ofthese angles create a pair of angles that are complementary. Exterior Angles of Triangles Exercises 1. tanT 2. 3; 2 3. X1T = Extend The sum of the remote interior angles of any convex polygon is equal to E + (n - 3)180 where E is the measure of the exterior angle and n is the number of sides of thepolygon. Relationships in Triangles Activity 41 Investigate 1. They are congruent. 2. The triangles are congruent by SAS. 3. BE ⬵ AD by CPCTC Answers 115 (continued) Extend 4. They are congruent. 5. 䉭FAC ⬵ 䉭BAE by SAS so CF ⬵ BE by CPCTC Investigate Further 6. 7. 8. 9. they are equal to120 Each angle of the triangle must be less than 120. yes It is equal to the sum of the lengths of the two shortest sides of the triangle. Isosceles Triangles Activity 42 Investigate 1. isosceles 2. ⬔DBC and ⬔DCB are congruent therefore 䉭BCD *remains ) isosceles 3. AD is the perpendicular bisector of BCInvestigate 4. The area of 䉭ABC is twice the area of ADFE. 5. the height is half that of 䉭ABC. 6. DE = 1 BC. 2 7. Each of the smaller triangles, 䉭ADE and 䉭DEF, have base and height that are half those of 䉭ABC. Therefore, the areas of each smaller triangle is 1 ? 1 5 1 䉭ABC. Together, the areas of䉭ADE and 2 2 4 䉭DEF must be half the area of 䉭ABC. 8. They have the same area. 9. The slopes of 䉭DEF are equal to the corresponding slopes of 䉭ABC. The measures of the sides of 䉭DEF are one-half the measure of the corresponding sides of 䉭ABC. This is a demonstration of the midsegmenttheorem. 10. yes Orthocenters Activity 43 Investigate 1. a. when the triangle is acute; b. when the triangle is right; c. when the triangle is obtuse 2. the orthocenter of 䉭ABD is C; the orthocenter of 䉭BCD is A; the orthocenter of 䉭ACD is B Investigate 3. The circumferences and areas of all four circles areequal. Quadrilaterals Activity 44 Investigate Using Different Menus to Construct Special Quadrilaterals Activity 45 Investigate 1–2. kite; two pairs of congruent, adjacent sides Investigate Further 3. 4. 5. 6. rectangle Check students’ work. rhombus rhombus Extend rectangle; congruent diagonals bisect eachother Parallelograms and Triangles Activity 46 Investigate 1. Check students’ work. 2. ⬔BAD is twice ⬔AFD; ⬔ABC is twice ⬔BEC 3. ⬔BAD = ⬔AFD + ⬔ADF by the Exterior Angle Theorem; ⬔AFD ⬵ ⬔ADF by the 3. Isosceles Triangle Theorem; therefore, by substitution, ⬔BAD = 2⬔AFD or 1 ⬔BAD= ⬔AFD; same rea2 soning used to show 1 ⬔ABC = ⬔BEC 2 4. ⬔BEC + ⬔AFD = 90, therefore ⬔CHD = 90 5. ⬔HCD ⬵ ⬔BEC and ⬔CDH ⬵ ⬔AFD 6. Corresponding Angles Postulate Regular Polygons Activity 47 Investigate 1. parallelogram 2. They are equal; They are parallel to BD. 3. One pair ofopposite sides of EHGF are congruent and parallel. 116 4. parallelogram 5. rectangle 6. The diagonals of a rhombus are perpendicular, therefore the sides of EHGF are perpendicular. 7. rhombus 8. square 9. rhombus 10. rectangle 11. If a quadrilateral has congruent diagonals, then the figure formed byjoining consecutive midpoints is a rhombus. 12. If a quadrilateral has perpendicular diagonals, then the figure formed by joining consecutive midpoints is a rectangle. 13. Opposite sides of a kite are not parallel. Both pairs of opposite sides of an isosceles trapezoid are not parallel. Answers 1. 2. 3. 4.rhombus 108 isosceles 36 Technology Activities © Pearson Education, Inc. All rights reserved. Answers Answers (continued) 5. m⬔GDE = m⬔GFE = 72; m⬔E = 108 6. 108, since the measures of a quadrilateral have a sum of 360 7. DE ⬵ EF 8. Since opposite angles are congruent, DGFE is aparallelogram. Because a pair of adjacent sides is congruent, DGFE is a rhombus. (n 2 1 1) (n2 2 1) n 2 2 3 = (A2 ? A2 – 1)/2 = A2 = (A2 ? A2 + 1)/2 = A2 + 2 = (A3 ? A3 – 1)/2 = A3 = (A3 ? A3 + 1)/2 = A3 + 2 = (A4 ? A4 – 1)/2 = A4 = (A4 ? A4 + 1)/2 n which produces the following triples Extend 9. 10. 11.12. 2. Pythagoras’ method uses the formulas rhombus Check students’ work. Check students’ work. Draw the first diagonal such that it is parallel to one of the sides of the regular polygon. The second diagonal will join vertices that are adjacent to the first diagonal such that the diagonals intersect. 3 5 7Investigate area of AECF = = areas are equal Check students’ work. E and G are midpoints of BC and CD, respectively; F coincides with C Using x to represent the length of one side of ABCD, 2 each triangle has measure 1 2 ? x ? 2x = x . The areas are equal. The sides to which the segments are drawnare divided into segments in a ratio of 2:2:1. The sides to which the segments are drawn are divided into segments in a ratio of 1:1:1. The sides to which the segments are drawn are divided into equal segments. The sides to which the segments are drawn are divided into segments in a ratio of 2:2: . . .:2:1. 9x2 3. 4. 5. © Pearson Education, Inc. All rights reserved. 6. 7. 8. 9. 3x2 3x2 Pythagorean Triples Investigate 1. Plato’s method uses the formulas n2 - 1 4 4 = A2 ? A2/4 – 1 = A2 + 2 = A3 ? A3/4 – 1 = A3 + 2 = A4 ? A4/4 – 1 n n = A2 = A3 = A4 n2 + 1 4 = A2 ? A2/4 + 1 = A3 ? A3/4 + 1 = A4 ? A4/4 +1 which produces the following triples n 4 6 n2 - 1 4 3 8 Technology Activities n 4 6 2n + 1 2n2 + 2n 2n2 + 2n + 1 produce the following triples n 1 2 3 3x2; the Activity 49 3 5 7 (n 2 1 1) 2 5 13 25 1 = 2?A2+1 = 2?A2?A2+2?A2 = 2?A2?A2+2?A2+1 = A2+1 = 2?A3+1 = 2?A3?A3+2?A3 = 2?A3?A3+2?A3+1= A3+1 = 2?A4+1 = 2?A4?A4+2?A4 = 2?A4?A4+2?A4+1 Activity 48 1. both ratios are 2:1 2. the area of ABCD is 3x ? 3x = 9x2; the area of 2 䉭ABE = area 䉭ADF = 1 2 ? 2x ? 3x = 3x ; the n 3. The formulas n Area (n2 2 1) 2 4 12 24 n 2n + 1 3 5 7 2n2 + 2n 4 12 24 2n2 + 2n + 1 5 13 25 4. Plato’s methodand Pythagoras’ method generate triples from the same ‘family’ of integers, but not necessarily the exact same numbers. For example, the Pythagoras method produces the triple 5, 12, 13 whereas Plato’s method produces 10, 24, 26, which is a multiple of 5, 12, 13. The unattributed method produces thesame triples as the Pythagoras method. Similarity Activity 50 Investigate 1. parallelogram 2. The sides of PQRS are midsegments, and therefore parallel, to the sides of ABCD. 3. corresponding angles are congruent; corresponding sides are in proportion 4. 2:1; 4:1 5. yes Extend 6. corresponding sides arein a ratio of 2:1; corresponding angles are congruent 7. the ratio of the areas is 4:1 8. Check student’s work. Investigate Further n2 + 1 4 5 10 9. ILMJ ⬃ KOPN; KIJN ⬃ OLMP 10. Answers may vary. Sample: IL : KO (for ILMJ ⬃ KOPN); KI : OL (for KIJN ⬃ OLMP) 11. Check students’ work. Answers 117Answers (continued) Volume Activity 51 Investigate 1. 12 - 4x 2. V(x) = x ? x ? (12 - 4x) = 12x2 - 4x3 Rotations Investigate 1. 2. 3. 4. 3. 2 4. 16 5. S(x) = 2 ? x ? x + 4 ? x ? (12 - 4x) = -14x2 + 48x 6. 1.714 7. no 8. (2.606, 30); maximum volume 10.695 each angle = 120; hexagon each angle 90; square 100is not a factor of 360 Any number that is a factor of 360 can used with a whole number amount of rotations to produce a regular polygon. Extend 5. It is supplementary. 6. Yes 7. The angles that have a vertex at A are equal to the size of rotation. Any other pair of angles formed will be supplementary to theangle of rotation. 8. A square can be used to form a tessellation. 9. Their angles must be factors of 360. 10. The sum of the angles around any point equals 360. 11. Regular pentagons and rhombuses can be used to form a semi-pure tessellation. Extend 9. V(x) = px2(12 - 2px) where x is the radius of thecontainer 10. r = 1.273 and h = 4; 20.372 11. S(x) = 2px2 + 2px(12 - 2px) 12. r = 1.136 and h = 4.864 13. The shape of a package, in addition to its weight, is an important consideration—Can the package be stacked on other packages? Can it fit into certain spaces on a truck or plane? Is it difficult tocarry? etc. . . . Web–Based Activities Activity 52 Algebra 1 Using the Web to Study Mathematics Check students’ work. Mining Data Found on the Internet year Time 52 45.9 56 44.5 60 44.5 64 43.6 68 42.8 72 42.81 76 42.55 80 41.6 84 41.65 88 41.98 92 42.11 Step 2 and 3: There is a moderate linearrelationship. Line of best fit: y = -0.0950681818x + 230.5653636 Step 4: Actual 1996 Winning time: 41.95; Line of best fit prediction = 40.80927273; The data appears to be leveling off from 1980–1992. This would seem reasonable since winning times cannot continue to decrease at the same rate. 118Answers Technology Activities © Pearson Education, Inc. All rights reserved. Check students’ work. Step 1: Winning Times for the Olympics Women’s 4 3 100–meter Relay Answers (continued) Generating Quizzes on the Web Check students’ work. Graphing the Solar System Investigate 1. 9.538 2.0.5341 2 x2 1 y 51 2 9.538 9.5232 (x 1 0.5341) 2 y2 4. 1 51 2 9.538 9.5232 5. 3. b = 9.523; 6. An eccentricity close to zero, such as 0.056, gives the appearance of a circle. Notice that the major axis, 2 ? 9.538 = 19.076, is slightly larger than the minor axis, 2 ? 9.523 = 19.046. © Pearson Education, Inc.All rights reserved. Exercises 1. The average distances to the Sun for the terrestrial planets are much closer to the Sun than the Jovian planets. A scale drawing of all nine planets would make it difficult to see the terrestrial planets. 2. Answers may vary: Pluto’s average distance to the Sun is about 100times greater than Mercury’s average distance to the Sun. A sketch would need to be about 25 feet wide in order to reasonably see the orbits of all nine planets. y2 x2 3. 1 51 2 39.785 38.5422 4. Pluto’s orbit is much more elliptical (e = 0.248) than Neptune’s (which is nearly circular with e = 0.009).Because of this, there are times when Pluto’s orbit brings the planet closer to the sun than Neptune’s. 5. The earth is tilted more towards the Sun during the summer months, thereby creating a warmer climate. 6. There would be extreme temperature changes if the Earth’s orbit was more elliptical. Buildinga Web Page About Buildings Check students’ work. Creating Dynamic Images for the Web Check students’ work. Technology Activities Answers 119 molecular geometry practice worksheet answers. lesson 8.1 practice a geometry worksheet answers. lesson 8.2 practice a geometry worksheet answers.lesson 3.4 practice a geometry worksheet answers. lesson 3.5 practice a geometry worksheet answers. lesson 8.1 practice b geometry worksheet answers. geometry conditional statement practice worksheet answers. lesson 3.1 practice a geometry worksheet answers

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